Trigonometry class 10 solved Archives – Mathemafia https://mathemafia.com/tag/trigonometry-class-10-solved/ Mathematics is like music - Enjoy it!!! Thu, 30 Oct 2025 04:21:19 +0000 en-US hourly 1 https://wordpress.org/?v=7.0 https://mathemafia.com/wp-content/uploads/2020/08/cropped-M__Logo__SM-1-32x32.png Trigonometry class 10 solved Archives – Mathemafia https://mathemafia.com/tag/trigonometry-class-10-solved/ 32 32 Exercise 8.3 (NCERT) – Introduction to Trigonometry, Class 10 – All Solutions https://mathemafia.com/exercise-8-3-ncert-introduction-to-trigonometry-class-10-all-solutions/ Mon, 17 Feb 2025 11:47:05 +0000 https://mathemafia.com/?p=11867 Complete solution of Exercise 8.3, Chapter 8, Introduction to Trigonometry of NCERT book of Mathematics for Class 10. All questions are based on the trigonometric ratios and of "Prove that" kind. The solutions are written in a very professional manner considering the level of understanding of the class 10 students. Every solution includes proper notes, labels and recommendations, where ever required. All trigonometric equations are written in proper Mathematics syntax. Full responsive page makes the tutorials easy to read on any device.

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Exercise 8.3 (NCERT) - Trigonometry, Class 10

Chapter 8 Exercise 8.3 Class 10

Below are the Quick links for all questions of Exercise 8.3, Trigonometry, Class 10.

Click a link to view the solution of corresponding question.

3 Idiots 😜 Identities – Don’t forget

(1)  π‘ π‘–𝑛2𝐴+π‘π‘œπ‘ 2𝐴=1(2)  1+π‘‘π‘Žπ‘›2𝐴=𝑠𝑒𝑐2𝐴(3)  1+π‘π‘œπ‘‘2𝐴=π‘π‘œπ‘ π‘’π‘2𝐴

Entire exercise is based on the 3 identities mentioned above. In case you forget the last 2 identities, you can derive them quickly from the first identity. Watch the video embedded below to know how.Β 

Below are the solutions of every question of exercise 8.3Β of chapter 8, Introduction toΒ Trigonometry, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.Β Β 

Question 1

Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Tip: tan A is reciprocal of cot A. So solve it first.Β 

π‘‘π‘Žπ‘›π΄=1π‘π‘œπ‘‘π΄

sec A

Let us find sec A now because we know the identity between sec A and tan A and we already know the value of tan A in terms of cot A.

𝑠𝑒𝑐2𝐴⇒𝑠𝑒𝑐𝐴=1+π‘‘π‘Žπ‘›2𝐴=1+1π‘π‘œπ‘‘2𝐴=1+π‘π‘œπ‘‘2π΄π‘π‘œπ‘‘2𝐴=1+π‘π‘œπ‘‘2π΄π‘π‘œπ‘‘2π΄β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš

sin A

Now we will use the value of sec A to find the value of sin A in terms of cot A. We know thatΒ 

π‘π‘œπ‘ π΄π‘π‘œπ‘ π΄=1𝑠𝑒𝑐𝐴=π‘π‘œπ‘‘2𝐴1+π‘π‘œπ‘‘2π΄β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš

Use the value of cos A in the following identity

𝑠𝑖𝑛2𝐴+π‘π‘œπ‘ 2𝐴=1
⇒𝑠𝑖𝑛2𝐴=1βˆ’π‘π‘œπ‘ 2𝐴=1–(π‘π‘œπ‘‘2𝐴1+π‘π‘œπ‘‘2π΄β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš)2=1β€“π‘π‘œπ‘‘2𝐴1+π‘π‘œπ‘‘2𝐴=1+π‘π‘œπ‘‘2π΄β€“π‘π‘œπ‘‘2𝐴1+π‘π‘œπ‘‘2𝐴=11+π‘π‘œπ‘‘2𝐴

Question 2

Write all the other trigonometric ratios of ∠A in terms of sec A.

Tip: cos A is reciprocal of sec A. So solve it first.Β 

π‘π‘œπ‘ π΄=1𝑠𝑒𝑐𝐴

sin A

We know that

β‡’  β‡’  π‘ π‘–𝑛2𝐴+π‘π‘œπ‘ 2𝐴=1𝑠𝑖𝑛2𝐴=1β€“π‘π‘œπ‘ 2𝐴=1–1𝑠𝑒𝑐2𝐴=𝑠𝑒𝑐2π΄βˆ’1𝑠𝑒𝑐2𝐴𝑠𝑖𝑛𝐴=𝑠𝑒𝑐2π΄βˆ’1𝑠𝑒𝑐2π΄β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=𝑠𝑒𝑐2π΄βˆ’1β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆšπ‘ π‘’π‘π΄

cosec A

Now cosec A is reciprocal of sin A, hence

π‘π‘œπ‘ π‘’π‘π΄=1𝑠𝑖𝑛𝐴=𝑠𝑒𝑐𝐴𝑠𝑒𝑐2π΄βˆ’1β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš

tan A

Method 1

Use Identity1+π‘‘π‘Žπ‘›2𝐴=𝑠𝑒𝑐2𝐴
β‡’  β‡’  π‘‘π‘Žπ‘›2𝐴=𝑠𝑒𝑐2𝐴–1π‘‘π‘Žπ‘›π΄=𝑠𝑒𝑐2𝐴–1β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš

Method 2

As we know the values of sin A and cos A, we can easily find tan A

π‘‘π‘Žπ‘›π΄=π‘ π‘–π‘›π΄π‘π‘œπ‘ π΄=𝑠𝑒𝑐2π΄βˆ’1βˆšπ‘ π‘’π‘π΄1𝑠𝑒𝑐𝐴=𝑠𝑒𝑐2π΄βˆ’1β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆšπ‘ π‘’π‘π΄Γ—π‘ π‘’π‘π΄=𝑠𝑒𝑐2π΄βˆ’1β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš

cot A

Now cot A is reciprocal of tan A, hence

π‘π‘œπ‘‘π΄=1π‘‘π‘Žπ‘›π΄=1𝑠𝑒𝑐2𝐴–1β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš

Question 3

Choose the correct option. Justify your choice.

(𝑖)  9  π‘ π‘’𝑐2𝐴  β€“  9  π‘‘π‘Žπ‘›2𝐴=(𝐴)  1  (𝐡)  9  (𝐢)  8  (𝐷)  0
=  9  π‘ π‘’𝑐2𝐴  β€“  9  π‘‘π‘Žπ‘›2𝐴  9Γ—(𝑠𝑒𝑐2𝐴  β€“  π‘‘π‘Žπ‘›2𝐴)
β‡’  We know1+π‘‘π‘Žπ‘›2𝐴=𝑠𝑒𝑐2𝐴𝑠𝑒𝑐2π΄β€“π‘‘π‘Žπ‘›2𝐴=1
==  9Γ—1  9

Answer: (B)

(𝑖𝑖)  (1+π‘‘π‘Žπ‘›ΞΈ+𝑠𝑒𝑐θ)(1+π‘π‘œπ‘‘ΞΈβ€“π‘π‘œπ‘ π‘’π‘ΞΈ)=(𝐴)  0  (𝐡)  1  (𝐢)  2  (𝐷)  β€“1
(1+π‘‘π‘Žπ‘›ΞΈ+𝑠𝑒𝑐θ)(1+π‘π‘œπ‘‘ΞΈβ€“π‘π‘œπ‘ π‘’π‘ΞΈ)=(1+π‘ π‘–π‘›ΞΈπ‘π‘œπ‘ ΞΈ+1π‘π‘œπ‘ ΞΈ)(1+π‘π‘œπ‘ ΞΈπ‘ π‘–π‘›ΞΈβ€“1𝑠𝑖𝑛θ)=(π‘π‘œπ‘ ΞΈ+𝑠𝑖𝑛θ+1π‘π‘œπ‘ ΞΈ)(𝑠𝑖𝑛θ+π‘π‘œπ‘ ΞΈβ€“1𝑠𝑖𝑛θ)=(𝑠𝑖𝑛θ+π‘π‘œπ‘ ΞΈ+1π‘π‘œπ‘ ΞΈ)(𝑠𝑖𝑛θ+π‘π‘œπ‘ ΞΈβ€“1𝑠𝑖𝑛θ)=(𝑠𝑖𝑛θ+π‘π‘œπ‘ ΞΈ)2–1𝑠𝑖𝑛θ  π‘π‘œπ‘ ΞΈ=𝑠𝑖𝑛2ΞΈ+π‘π‘œπ‘ 2ΞΈ+2 sinΞΈ  π‘π‘œπ‘ ΞΈβ€“1𝑠𝑖𝑛θ  π‘π‘œπ‘ ΞΈ=1+2 sinΞΈ  π‘π‘œπ‘ ΞΈβ€“1𝑠𝑖𝑛θ  π‘π‘œπ‘ ΞΈ=2 sinΞΈ  π‘π‘œπ‘ ΞΈπ‘ π‘–𝑛θ  π‘π‘œπ‘ ΞΈ=2

Answer: (C)

(𝑖𝑖𝑖)  (𝑠𝑒𝑐𝐴+π‘‘π‘Žπ‘›π΄)(1–𝑠𝑖𝑛𝐴)=(𝐴)  π‘ π‘’𝑐𝐴  (𝐡)  π‘ π‘–𝑛𝐴(𝐢)  π‘π‘œπ‘ π‘’𝑐𝐴  (𝐷)  π‘π‘œπ‘ π΄

Let us convert all terms to sin and cos

(𝑠𝑒𝑐𝐴+π‘‘π‘Žπ‘›π΄)(1  β€“  π‘ π‘–𝑛𝐴)=(1π‘π‘œπ‘ π΄+π‘ π‘–π‘›π΄π‘π‘œπ‘ π΄)(1  β€“  π‘ π‘–𝑛𝐴)=(1+π‘ π‘–π‘›π΄π‘π‘œπ‘ π΄)(1  β€“  π‘ π‘–𝑛𝐴)=1–𝑠𝑖𝑛2π΄π‘π‘œπ‘ π΄=π‘π‘œπ‘ 2π΄π‘π‘œπ‘ π΄=π‘π‘œπ‘ π΄

Answer: (D)

(𝑖𝑣)  1+π‘‘π‘Žπ‘›2𝐴1+π‘π‘œπ‘‘2𝐴=(𝐴)  π‘ π‘’𝑐2𝐴  (𝐡)  βˆ’1(𝐢)  π‘π‘œπ‘‘2𝐴  (𝐷) tan2𝐴

Method 1:

1+π‘‘π‘Žπ‘›2𝐴1+π‘π‘œπ‘‘2𝐴=1+π‘‘π‘Žπ‘›2𝐴1+1π‘‘π‘Žπ‘›2𝐴=1+π‘‘π‘Žπ‘›2π΄π‘‘π‘Žπ‘›2𝐴+1π‘‘π‘Žπ‘›2𝐴=(1+π‘‘π‘Žπ‘›2𝐴)Γ—π‘‘π‘Žπ‘›2𝐴(1+π‘‘π‘Žπ‘›2𝐴)=π‘‘π‘Žπ‘›2𝐴

Method 2:

In case you are not able to recall the first method, you can convert everything to sin and cos to solve it.Β 

1+π‘‘π‘Žπ‘›2𝐴1+π‘π‘œπ‘‘2𝐴=1+𝑠𝑖𝑛2π΄π‘π‘œπ‘ 2𝐴1+π‘π‘œπ‘ 2𝐴𝑠𝑖𝑛2𝐴=π‘π‘œπ‘ 2𝐴+𝑠𝑖𝑛2π΄π‘π‘œπ‘ 2𝐴𝑠𝑖𝑛2𝐴+π‘π‘œπ‘ 2𝐴𝑠𝑖𝑛2𝐴=1π‘π‘œπ‘ 2𝐴1𝑠𝑖𝑛2𝐴=1π‘π‘œπ‘ 2𝐴×𝑠𝑖𝑛2𝐴=π‘‘π‘Žπ‘›2𝐴

Answer: (D)

Question 4

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(𝑖)  (π‘π‘œπ‘ π‘’π‘ΞΈβ€“π‘π‘œπ‘‘ΞΈ)2=1βˆ’π‘π‘œπ‘ ΞΈ1+π‘π‘œπ‘ ΞΈ
𝐿𝐻𝑆=(π‘π‘œπ‘ π‘’π‘ΞΈβ€“π‘π‘œπ‘‘ΞΈ)2=(1π‘ π‘–π‘›ΞΈβ€“π‘π‘œπ‘ ΞΈπ‘ π‘–π‘›ΞΈ)2=(1βˆ’π‘π‘œπ‘ ΞΈπ‘ π‘–π‘›ΞΈ)2=(1βˆ’π‘π‘œπ‘ ΞΈ)2𝑠𝑖𝑛2ΞΈ=(1βˆ’π‘π‘œπ‘ ΞΈ)21βˆ’π‘π‘œπ‘ 2ΞΈ=(1βˆ’π‘π‘œπ‘ ΞΈ)2(1βˆ’π‘π‘œπ‘ ΞΈ)(1+π‘π‘œπ‘ ΞΈ)=1βˆ’π‘π‘œπ‘ ΞΈ1+π‘π‘œπ‘ ΞΈ=𝑅𝐻𝑆
(𝑖𝑖)  π‘π‘œπ‘ π΄1+𝑠𝑖𝑛𝐴+1+π‘ π‘–π‘›π΄π‘π‘œπ‘ π΄=2  π‘ π‘’𝑐𝐴
𝐿𝐻𝑆=π‘π‘œπ‘ π΄1+𝑠𝑖𝑛𝐴+1+π‘ π‘–π‘›π΄π‘π‘œπ‘ π΄=π‘π‘œπ‘ 2𝐴+(1+𝑠𝑖𝑛𝐴)2(1+𝑠𝑖𝑛𝐴)(π‘π‘œπ‘ π΄)=π‘π‘œπ‘ 2𝐴+1+𝑠𝑖𝑛2𝐴+2𝑠𝑖𝑛𝐴(1+𝑠𝑖𝑛𝐴)(π‘π‘œπ‘ π΄)=𝑠𝑖𝑛2𝐴+π‘π‘œπ‘ 2𝐴+1+2𝑠𝑖𝑛𝐴(1+𝑠𝑖𝑛𝐴)(π‘π‘œπ‘ π΄)=1+1+2𝑠𝑖𝑛𝐴(1+𝑠𝑖𝑛𝐴)(π‘π‘œπ‘ π΄)=2+2𝑠𝑖𝑛𝐴(1+𝑠𝑖𝑛𝐴)(π‘π‘œπ‘ π΄)=2(1+𝑠𝑖𝑛𝐴)(1+𝑠𝑖𝑛𝐴)(π‘π‘œπ‘ π΄)=2π‘π‘œπ‘ π΄=2𝑠𝑒𝑐𝐴=𝑅𝐻𝑆
(𝑖𝑖𝑖)  π‘‘π‘Žπ‘›ΞΈ1βˆ’π‘π‘œπ‘‘ΞΈ+π‘π‘œπ‘‘ΞΈ1βˆ’π‘‘π‘Žπ‘›ΞΈ=1+𝑠𝑒𝑐θ  π‘π‘œπ‘ π‘’π‘ΞΈ

[Hint : Write the expression in terms of sin ΞΈ and cos ΞΈ]

𝐿𝐻𝑆=π‘‘π‘Žπ‘›ΞΈ1βˆ’π‘π‘œπ‘‘ΞΈ+π‘π‘œπ‘‘ΞΈ1βˆ’π‘‘π‘Žπ‘›ΞΈ=π‘ π‘–π‘›ΞΈπ‘π‘œπ‘ ΞΈ1βˆ’π‘π‘œπ‘ ΞΈπ‘ π‘–π‘›ΞΈ+π‘π‘œπ‘ ΞΈπ‘ π‘–π‘›ΞΈ1βˆ’π‘ π‘–π‘›ΞΈπ‘π‘œπ‘ ΞΈ=π‘ π‘–π‘›ΞΈπ‘π‘œπ‘ ΞΈπ‘ π‘–π‘›ΞΈβ€“π‘π‘œπ‘ ΞΈπ‘ π‘–π‘›ΞΈ+π‘π‘œπ‘ ΞΈπ‘ π‘–π‘›ΞΈπ‘π‘œπ‘ ΞΈβˆ’π‘ π‘–π‘›ΞΈπ‘π‘œπ‘ ΞΈ=π‘ π‘–π‘›ΞΈπ‘π‘œπ‘ ΞΈΓ—π‘ π‘–π‘›ΞΈπ‘ π‘–π‘›ΞΈβ€“π‘π‘œπ‘ ΞΈ+π‘π‘œπ‘ ΞΈπ‘ π‘–π‘›ΞΈΓ—π‘π‘œπ‘ ΞΈπ‘π‘œπ‘ ΞΈβˆ’π‘ π‘–π‘›ΞΈ=𝑠𝑖𝑛2ΞΈπ‘π‘œπ‘ ΞΈΓ—(π‘ π‘–π‘›ΞΈβ€“π‘π‘œπ‘ ΞΈ)+π‘π‘œπ‘ 2θ𝑠𝑖𝑛θ×(π‘π‘œπ‘ ΞΈβˆ’π‘ π‘–π‘›ΞΈ)=𝑠𝑖𝑛2ΞΈπ‘π‘œπ‘ ΞΈΓ—(π‘ π‘–π‘›ΞΈβ€“π‘π‘œπ‘ ΞΈ)β€“π‘π‘œπ‘ 2θ𝑠𝑖𝑛θ×(π‘ π‘–π‘›ΞΈβˆ’π‘π‘œπ‘ ΞΈ)=𝑠𝑖𝑛3ΞΈβ€“π‘π‘œπ‘ 3ΞΈπ‘π‘œπ‘ ΞΈΓ—π‘ π‘–π‘›ΞΈΓ—(π‘ π‘–π‘›ΞΈβ€“π‘π‘œπ‘ ΞΈ)=(π‘ π‘–π‘›ΞΈβ€“π‘π‘œπ‘ ΞΈ)Γ—(𝑠𝑖𝑛2ΞΈ+π‘π‘œπ‘ 2ΞΈ+π‘ π‘–π‘›ΞΈΓ—π‘π‘œπ‘ ΞΈ)π‘π‘œπ‘ ΞΈΓ—π‘ π‘–π‘›ΞΈΓ—(π‘ π‘–π‘›ΞΈβ€“π‘π‘œπ‘ ΞΈ)=1+π‘ π‘–π‘›ΞΈΓ—π‘π‘œπ‘ ΞΈπ‘π‘œπ‘ ΞΈΓ—π‘ π‘–π‘›ΞΈ=1π‘π‘œπ‘ ΞΈΓ—π‘ π‘–π‘›ΞΈ+π‘ π‘–π‘›ΞΈΓ—π‘π‘œπ‘ ΞΈπ‘π‘œπ‘ ΞΈΓ—π‘ π‘–π‘›ΞΈ=π‘ π‘’π‘ΞΈΓ—π‘π‘œπ‘ π‘’π‘ΞΈ+1=1+π‘ π‘’π‘ΞΈΓ—π‘π‘œπ‘ π‘’π‘ΞΈ=𝑅𝐻𝑆
(𝑖𝑣)  1+𝑠𝑒𝑐𝐴𝑠𝑒𝑐𝐴=𝑠𝑖𝑛2𝐴1βˆ’π‘π‘œπ‘ π΄

[Hint : Simplify LHS and RHS separately]

𝐿𝐻𝑆=1+𝑠𝑒𝑐𝐴𝑠𝑒𝑐𝐴=1+1π‘π‘œπ‘ π΄1π‘π‘œπ‘ π΄=π‘π‘œπ‘ π΄+1π‘π‘œπ‘ π΄1π‘π‘œπ‘ π΄=π‘π‘œπ‘ π΄+1π‘π‘œπ‘ π΄Γ—π‘π‘œπ‘ π΄1=1+π‘π‘œπ‘ π΄  β€¦  β€¦(1)
𝑅𝐻𝑆=𝑠𝑖𝑛2𝐴1βˆ’π‘π‘œπ‘ π΄=1βˆ’π‘π‘œπ‘ 2𝐴1βˆ’π‘π‘œπ‘ π΄=(1βˆ’π‘π‘œπ‘ π΄)(1+π‘π‘œπ‘ π΄)1βˆ’π‘π‘œπ‘ π΄=1+π‘π‘œπ‘ π΄  β€¦  β€¦(2)

From (1) and (2)

LHS = RHS

(𝑣)  π‘π‘œπ‘ π΄β€“𝑠𝑖𝑛𝐴+1π‘π‘œπ‘ π΄+𝑠𝑖𝑛𝐴–1=π‘π‘œπ‘ π‘’π‘π΄+π‘π‘œπ‘‘π΄using the identityπ‘π‘œπ‘ π‘’π‘2𝐴=1+π‘π‘œπ‘‘2𝐴

[Hint : Write the expression in terms of sin ΞΈ and cos ΞΈ]

𝐿𝐻𝑆=π‘π‘œπ‘ π΄β€“π‘ π‘–π‘›π΄+1π‘π‘œπ‘ π΄+𝑠𝑖𝑛𝐴–1

As we are asked to use the identity

π‘π‘œπ‘ π‘’π‘2𝐴=1+π‘π‘œπ‘‘2𝐴

Divide both numerator and denominator by sin A to convert the terms in cot and cosec.

𝐿𝐻𝑆=π‘π‘œπ‘ π΄β€“π‘ π‘–π‘›π΄+1π‘ π‘–π‘›π΄π‘π‘œπ‘ π΄+𝑠𝑖𝑛𝐴–1𝑠𝑖𝑛𝐴=π‘π‘œπ‘‘π΄β€“1+π‘π‘œπ‘ π‘’π‘π΄π‘π‘œπ‘‘π΄+1β€“π‘π‘œπ‘ π‘’π‘π΄=π‘π‘œπ‘ π‘’π‘π΄+π‘π‘œπ‘‘π΄β€“1π‘π‘œπ‘‘π΄+1β€“π‘π‘œπ‘ π‘’π‘π΄  β€¦(1)
From identityπ‘π‘œπ‘ π‘’π‘2𝐴=1+π‘π‘œπ‘‘2𝐴We get1=π‘π‘œπ‘ π‘’π‘2π΄β€“π‘π‘œπ‘‘2𝐴Let us use it (1)
𝐿𝐻𝑆=(π‘π‘œπ‘ π‘’π‘π΄+π‘π‘œπ‘‘π΄)–(π‘π‘œπ‘ π‘’π‘2π΄βˆ’π‘π‘œπ‘‘2𝐴)π‘π‘œπ‘‘π΄+1β€“π‘π‘œπ‘ π‘’π‘π΄=(π‘π‘œπ‘ π‘’π‘π΄+π‘π‘œπ‘‘π΄)–(π‘π‘œπ‘ π‘’π‘π΄βˆ’π‘π‘œπ‘‘π΄)(π‘π‘œπ‘ π‘’π‘π΄+π‘π‘œπ‘‘π΄)1+π‘π‘œπ‘‘π΄β€“π‘π‘œπ‘ π‘’π‘π΄=(π‘π‘œπ‘ π‘’π‘π΄+π‘π‘œπ‘‘π΄)[1βˆ’(π‘π‘œπ‘ π‘’π‘π΄βˆ’π‘π‘œπ‘‘π΄)]1+π‘π‘œπ‘‘π΄β€“π‘π‘œπ‘ π‘’π‘π΄=(π‘π‘œπ‘ π‘’π‘π΄+π‘π‘œπ‘‘π΄)(1βˆ’π‘π‘œπ‘ π‘’π‘π΄+π‘π‘œπ‘‘π΄)1+π‘π‘œπ‘‘π΄β€“π‘π‘œπ‘ π‘’π‘π΄=(π‘π‘œπ‘ π‘’π‘π΄+π‘π‘œπ‘‘π΄)(1+π‘π‘œπ‘‘π΄βˆ’π‘π‘œπ‘ π‘’π‘π΄)1+π‘π‘œπ‘‘π΄β€“π‘π‘œπ‘ π‘’π‘π΄=π‘π‘œπ‘ π‘’π‘π΄+π‘π‘œπ‘‘π΄=𝑅𝐻𝑆
(𝑣𝑖)  1+𝑠𝑖𝑛𝐴1βˆ’π‘ π‘–π‘›π΄β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=𝑠𝑒𝑐𝐴+π‘‘π‘Žπ‘›π΄

IMP: LHS is more complicated because of square root, hence we will start from LHS.

𝐿𝐻𝑆=1+𝑠𝑖𝑛𝐴1βˆ’π‘ π‘–π‘›π΄β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=1+𝑠𝑖𝑛𝐴1βˆ’π‘ π‘–π‘›π΄Γ—1+𝑠𝑖𝑛𝐴1+π‘ π‘–π‘›π΄β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=(1+𝑠𝑖𝑛𝐴)21βˆ’π‘ π‘–π‘›2π΄β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=(1+𝑠𝑖𝑛𝐴)2π‘π‘œπ‘ 2π΄β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=1+π‘ π‘–π‘›π΄π‘π‘œπ‘ π΄=1π‘π‘œπ‘ π΄+π‘ π‘–π‘›π΄π‘π‘œπ‘ π΄=𝑠𝑒𝑐𝐴+π‘‘π‘Žπ‘›π΄=𝑅𝐻𝑆
(𝑣𝑖𝑖)  π‘ π‘–π‘›ΞΈβˆ’2𝑠𝑖𝑛3ΞΈ2π‘π‘œπ‘ 3ΞΈβˆ’π‘π‘œπ‘ ΞΈ=π‘‘π‘Žπ‘›ΞΈ
𝐿𝐻𝑆=π‘ π‘–π‘›ΞΈβˆ’2  π‘ π‘–𝑛3ΞΈ2  π‘π‘œπ‘ 3ΞΈβˆ’π‘π‘œπ‘ ΞΈ=𝑠𝑖𝑛θ  (1βˆ’2  π‘ π‘–𝑛2ΞΈ)π‘π‘œπ‘ ΞΈ  (2  π‘π‘œπ‘ 2ΞΈβˆ’1)=𝑠𝑖𝑛θ  (𝑠𝑖𝑛2ΞΈ+π‘π‘œπ‘ 2ΞΈβˆ’2  π‘ π‘–𝑛2ΞΈ)π‘π‘œπ‘ ΞΈ  [2  π‘π‘œπ‘ 2ΞΈβˆ’(𝑠𝑖𝑛2ΞΈ+π‘π‘œπ‘ 2ΞΈ)]=𝑠𝑖𝑛θ  (π‘π‘œπ‘ 2ΞΈβˆ’π‘ π‘–π‘›2ΞΈ)π‘π‘œπ‘ ΞΈ  (π‘π‘œπ‘ 2ΞΈβˆ’π‘ π‘–π‘›2ΞΈ)=π‘ π‘–π‘›ΞΈπ‘π‘œπ‘ ΞΈ=π‘‘π‘Žπ‘›ΞΈ=𝑅𝐻𝑆
(𝑣𝑖𝑖𝑖)  (𝑠𝑖𝑛𝐴+π‘π‘œπ‘ π‘’π‘π΄)2+(π‘π‘œπ‘ π΄+𝑠𝑒𝑐𝐴)2=7+π‘‘π‘Žπ‘›2𝐴+π‘π‘œπ‘‘2𝐴
𝐿𝐻𝑆=(𝑠𝑖𝑛𝐴+π‘π‘œπ‘ π‘’π‘π΄)2+(π‘π‘œπ‘ π΄+𝑠𝑒𝑐𝐴)2=𝑠𝑖𝑛2𝐴+π‘π‘œπ‘ π‘’π‘2𝐴+2  π‘ π‘–𝑛𝐴  π‘π‘œπ‘ π‘’𝑐𝐴+π‘π‘œπ‘ 2𝐴+𝑠𝑒𝑐2𝐴+2  π‘π‘œπ‘ π΄  π‘ π‘’𝑐𝐴=𝑠𝑖𝑛2𝐴+π‘π‘œπ‘ 2𝐴+π‘π‘œπ‘ π‘’π‘2𝐴+2  π‘ π‘–𝑛𝐴  1𝑠𝑖𝑛𝐴+𝑠𝑒𝑐2𝐴+2  π‘π‘œπ‘ π΄  1π‘π‘œπ‘ π΄=1+π‘π‘œπ‘ π‘’π‘2𝐴+2+𝑠𝑒𝑐2𝐴+2=5+π‘π‘œπ‘ π‘’π‘2𝐴+𝑠𝑒𝑐2𝐴=5+1+π‘π‘œπ‘‘2𝐴+1+π‘‘π‘Žπ‘›2𝐴=7+π‘‘π‘Žπ‘›2𝐴+π‘π‘œπ‘‘2𝐴=𝑅𝐻𝑆
(𝑖π‘₯)  (π‘π‘œπ‘ π‘’π‘π΄β€“π‘ π‘–π‘›π΄)(π‘ π‘’π‘π΄β€“π‘π‘œπ‘ π΄)=1π‘‘π‘Žπ‘›π΄+π‘π‘œπ‘‘π΄

[Hint : Simplify LHS and RHS separately]

Convert LHS and RHS in terms of sin and cos.

𝐿𝐻𝑆=(π‘π‘œπ‘ π‘’π‘π΄β€“π‘ π‘–π‘›π΄)(π‘ π‘’π‘π΄β€“π‘π‘œπ‘ π΄)=(1𝑠𝑖𝑛𝐴–𝑠𝑖𝑛𝐴)(1π‘π‘œπ‘ π΄β€“π‘π‘œπ‘ π΄)=(1βˆ’π‘ π‘–π‘›2𝐴𝑠𝑖𝑛𝐴)(1βˆ’π‘π‘œπ‘ 2π΄π‘π‘œπ‘ π΄)=(π‘π‘œπ‘ 2𝐴𝑠𝑖𝑛𝐴)(𝑠𝑖𝑛2π΄π‘π‘œπ‘ π΄)=𝑠𝑖𝑛𝐴  π‘π‘œπ‘ π΄  β€¦  β€¦(1)
𝑅𝐻𝑆=1π‘‘π‘Žπ‘›π΄+π‘π‘œπ‘‘π΄=1π‘ π‘–π‘›π΄π‘π‘œπ‘ π΄+π‘π‘œπ‘ π΄π‘ π‘–π‘›π΄=1𝑠𝑖𝑛2𝐴+π‘π‘œπ‘ 2𝐴𝑠𝑖𝑛𝐴  π‘π‘œπ‘ π΄=11𝑠𝑖𝑛𝐴  π‘π‘œπ‘ π΄=𝑠𝑖𝑛𝐴  π‘π‘œπ‘ π΄  β€¦  β€¦(2)

Frim (1) and (2)

LHS = RHS

(π‘₯)  1+π‘‘π‘Žπ‘›2𝐴1+π‘π‘œπ‘‘2𝐴=(1βˆ’π‘‘π‘Žπ‘›π΄1βˆ’π‘π‘œπ‘‘π΄)2=π‘‘π‘Žπ‘›2𝐴

There are three terms that we need to prove equal to each other. Β We will first take LHS and then middle term (MT) and prove both equal to RHS. Hence all three terms will be equal to each other.

𝐿𝐻𝑆=1+π‘‘π‘Žπ‘›2𝐴1+π‘π‘œπ‘‘2𝐴=1+𝑠𝑖𝑛2π΄π‘π‘œπ‘ 2𝐴1+π‘π‘œπ‘ 2𝐴𝑠𝑖𝑛2𝐴=π‘π‘œπ‘ 2𝐴+𝑠𝑖𝑛2π΄π‘π‘œπ‘ 2𝐴𝑠𝑖𝑛2𝐴+π‘π‘œπ‘ 2𝐴𝑠𝑖𝑛2𝐴=1π‘π‘œπ‘ 2𝐴1𝑠𝑖𝑛2𝐴=1π‘π‘œπ‘ 2𝐴×𝑠𝑖𝑛2𝐴1=π‘‘π‘Žπ‘›2𝐴=𝑅𝐻𝑆
𝑀𝑇=(1βˆ’π‘‘π‘Žπ‘›π΄1βˆ’π‘π‘œπ‘‘π΄)2=(1βˆ’π‘ π‘–π‘›π΄π‘π‘œπ‘ π΄1βˆ’π‘π‘œπ‘ π΄π‘ π‘–π‘›π΄)2=(π‘π‘œπ‘ π΄β€“π‘ π‘–π‘›π΄π‘π‘œπ‘ π΄π‘ π‘–π‘›π΄βˆ’π‘π‘œπ‘ π΄π‘ π‘–π‘›π΄)2=[π‘π‘œπ‘ π΄β€“π‘ π‘–π‘›π΄π‘π‘œπ‘ π΄Γ—π‘ π‘–π‘›π΄π‘ π‘–π‘›π΄βˆ’π‘π‘œπ‘ π΄]2=[βˆ’(π‘ π‘–π‘›π΄βˆ’π‘π‘œπ‘ π΄)π‘π‘œπ‘ π΄Γ—π‘ π‘–π‘›π΄π‘ π‘–π‘›π΄βˆ’π‘π‘œπ‘ π΄]2=[βˆ’π‘ π‘–π‘›π΄π‘π‘œπ‘ π΄]2=(βˆ’π‘‘π‘Žπ‘›π΄)2=π‘‘π‘Žπ‘›2𝐴=𝑅𝐻𝑆

Recommended Tutorial

Here is a recommended tutorial that you may find helpful understanding the basic concepts of Trigonometry.

\(\begin{aligned}Β  \end{aligned}\)

The post Exercise 8.3 (NCERT) – Introduction to Trigonometry, Class 10 – All Solutions appeared first on Mathemafia.

]]> Exercise 8.2 (NCERT) – Introduction to Trigonometry, Class 10 – All Solutions https://mathemafia.com/exercise-8-2-ncert-introduction-to-trigonometry-class-10-all-solutions-ncert/ Thu, 13 Feb 2025 05:09:38 +0000 https://mathemafia.com/?p=11077 Complete solution of Exercise 8.2, Chapter 8, Introduction to Trigonometry of NCERT book of Mathematics for Class 10. This exercise is more about finding the standard values the trigonometric ratios and hence really easy to solve. The solutions are written in pure mathematical way with mobile-friendly view, fully responsive. The free YouTube video tutorial of every question is embedded after the written solution. This educational content has been created with utmost care to make it easily readable and hence understandable by the students of class 10, especially preparing for the final board exams.

The post Exercise 8.2 (NCERT) – Introduction to Trigonometry, Class 10 – All Solutions appeared first on Mathemafia.

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Exercise 8.2 (NCERT) - Trigonometry, Class 10

Chapter 8 Exercise 8.2 Class 10

Below are the Quick links for all questions of Exercise 8.2, Trigonometry, Class 10.

Click a link to view the solution of corresponding question.

Below are the solutions of every question of Exercise 8.2Β of Chapter 8, Introduction toΒ Trigonometry, NCERT of Class 10. YouTube video tutorial of every question is also embedded along with the written solution.Β Β 

This exercise is primarily about the standard values of trigonometric ratios and that’s why the questions are quite easy to solve.

Very Important: Before you start solving the questions of this exercise, it is very important that you remember the table displayed below.

Trigonometric ratios of standard angles

∠A 0° 30° 45° 60° 90°
sin A
0
\(\begin{aligned} \frac {1}{2}\end{aligned} \)
\(\begin{aligned} \frac {1}{\sqrt{2}}\end{aligned} \)
\(\begin{aligned} \frac {\sqrt{3}}{2}\end{aligned} \)
1
cos A
1
\(\begin{aligned} \frac {\sqrt{3}}{2}\end{aligned} \)
\(\begin{aligned} \frac {1}{\sqrt{2}}\end{aligned} \)
\(\begin{aligned} \frac {1}{2}\end{aligned} \)
0
tan A
0
\(\begin{aligned} \frac {1}{\sqrt {3}}\end{aligned} \)
1
\(\begin{aligned} \sqrt {3} \end{aligned} \)
N.D.
cosec A
N.D.
2
\(\begin{aligned} \sqrt {2} \end{aligned} \)
\(\begin{aligned} \frac {2}{\sqrt{3}}\end{aligned} \)
1
sec A
1
\(\begin{aligned} \frac {2}{\sqrt{3}}\end{aligned} \)
\(\begin{aligned} \sqrt{2}\end{aligned} \)
2
N.D.
cot A
N.D.
\(\begin{aligned} \sqrt{3}\end{aligned} \)
1
\(\begin{aligned} \frac {1}{\sqrt{3}}\end{aligned} \)
1

N.D. – Not Defined

How to quickly make this table in exams

Let’s start using these trigonometric ratios of angles 0Β°, 30Β°, 45Β°, 60Β° and 90Β° to solve the questions.Β 

Question 1

Evaluate the following :

(𝑖)  π‘ π‘–𝑛60Β°π‘π‘œπ‘ 30Β°+𝑠𝑖𝑛30Β°π‘π‘œπ‘ 60Β°
=3β€Ύβˆš2Γ—3β€Ύβˆš2+12Γ—12=34+14=1
(𝑖𝑖)  2π‘‘π‘Žπ‘›245Β°+π‘π‘œπ‘ 230°–𝑠𝑖𝑛260Β°
=2Γ—12+(3β€Ύβˆš2)2–(3β€Ύβˆš2)2=2+34–34=2
(𝑖𝑖𝑖)  π‘π‘œπ‘ 45°𝑠𝑒𝑐30Β°+π‘π‘œπ‘ π‘’π‘30Β°
=12√23√+2=12√2+23√3√=12β€ΎβˆšΓ—3β€Ύβˆš2+23β€Ύβˆš=3β€Ύβˆš22β€Ύβˆš+22β€Ύβˆš3β€Ύβˆš=3β€Ύβˆš22β€Ύβˆš+26β€Ύβˆš

Note: As per my understanding, this should be treated as the final answer, but in your book, the denominator has been rationalized. To achieve that, we need to further solve it as follows.Β 

Rationalizing the denominatorΒ 

=3β€Ύβˆš22β€Ύβˆš+26β€ΎβˆšΓ—22β€Ύβˆšβ€“26β€Ύβˆš22β€Ύβˆšβ€“26β€Ύβˆš=22β€Ύβˆš3β€Ύβˆšβ€“23β€Ύβˆš6β€Ύβˆš4(2β€Ύβˆš)2–4Γ—6=22β€Ύβˆš3β€Ύβˆšβ€“23β€Ύβˆš2β€Ύβˆš3β€Ύβˆš4Γ—2–4Γ—6=26β€Ύβˆšβ€“62β€Ύβˆš8βˆ’24=2(6β€Ύβˆšβ€“32β€Ύβˆš)βˆ’16=6β€Ύβˆšβ€“32β€Ύβˆšβˆ’8=32β€Ύβˆšβ€“6β€Ύβˆš8
(𝑖𝑣)  π‘ π‘–𝑛30Β°+π‘‘π‘Žπ‘›45Β°β€“π‘π‘œπ‘ π‘’π‘60°𝑠𝑒𝑐30Β°+π‘π‘œπ‘ 60Β°+π‘π‘œπ‘‘45Β°
=12+1–23√23√+12+1=3√+23βˆšβ€“423√4+3√+23√23√=33βˆšβ€“423√4+33√23√=33β€Ύβˆšβ€“423β€ΎβˆšΓ—23β€Ύβˆš4+33β€Ύβˆš=33β€Ύβˆšβ€“44+33β€Ύβˆš

Rationalizing the denominator

=33β€Ύβˆšβ€“44+33β€ΎβˆšΓ—4–33β€Ύβˆš4–33β€Ύβˆš=123β€Ύβˆšβ€“16–27+123β€Ύβˆš16βˆ’27=243β€Ύβˆšβ€“43βˆ’11=43–243β€Ύβˆš11
(𝑣)  5π‘π‘œπ‘ 260Β°+4𝑠𝑒𝑐230Β°β€“π‘‘π‘Žπ‘›245°𝑠𝑖𝑛230Β°+π‘π‘œπ‘ 230Β°
=5Γ—(12)2+4Γ—(23√)2βˆ’12(12)2+(3√2)2=5Γ—14+4Γ—43–114+34=54+163–144=15+64–1212=6712
Click the following video to watch it’s solution.

Question 2

Choose the correct option and justify your choice :

(𝑖)  2π‘‘π‘Žπ‘›30Β°1+π‘‘π‘Žπ‘›230Β°(𝐴)  π‘ π‘–𝑛60Β°  (𝐡)  π‘π‘œπ‘ 60Β°(𝐢)  π‘‘π‘Žπ‘›60Β°  (𝐷)  π‘ π‘–𝑛30Β°
2π‘‘π‘Žπ‘›30Β°1+π‘‘π‘Žπ‘›230Β°=2Γ—13√1+(13√)2=23√1+(13)=23√43=23β€ΎβˆšΓ—34=323β€Ύβˆš=3β€ΎβˆšΓ—3β€Ύβˆš23β€Ύβˆš=3β€Ύβˆš2=𝑠𝑖𝑛60Β°

Hence (A) is the answer.Β 

(𝑖𝑖)  1βˆ’π‘‘π‘Žπ‘›245Β°1+π‘‘π‘Žπ‘›245Β°(𝐴) tan90Β°  (𝐡)  1(𝐢)  π‘ π‘–𝑛45Β°  (𝐷)  0
=1βˆ’π‘‘π‘Žπ‘›245Β°1+π‘‘π‘Žπ‘›245Β°=1βˆ’121+12=02=0

Hence (D) is the answer.Β 

(iii) sin 2A = 2 sin A is true when A =Β 

(𝐴)  0Β°  (𝐡)  30Β°(𝐢)  45Β°  (𝐷)  60Β°

Let us take A = 0Β°

⇒𝑠𝑖𝑛𝐴⇒2π‘ π‘–π‘›π΄π΄π‘™π‘ π‘œ  π‘ π‘–𝑛2𝐴⇒2𝑠𝑖𝑛𝐴=𝑠𝑖𝑛0Β°=0=0=𝑠𝑖𝑛(2Γ—0)=𝑠𝑖𝑛0Β°=0=𝑠𝑖𝑛2𝐴

Hence (A) is the answer.

(𝑖𝑣)  2π‘‘π‘Žπ‘›30Β°1β€“π‘‘π‘Žπ‘›230Β°(𝐴)  π‘π‘œπ‘ 60Β°  (𝐡)  π‘ π‘–𝑛60Β°(𝐢)  π‘‘π‘Žπ‘›60Β°  (𝐷)  π‘ π‘–𝑛30Β°
2π‘‘π‘Žπ‘›30Β°1β€“π‘‘π‘Žπ‘›230Β°=2Γ—13√1–(13√)2=23√1–13=23√23=23β€ΎβˆšΓ—32=33β€Ύβˆš=3β€Ύβˆš=π‘‘π‘Žπ‘›60Β°

Hence (C) is the answer.

Click the following video to watch it’s solution.

Question 3

𝐼𝑓  π‘‘π‘Žπ‘›(𝐴+𝐡)=3β€Ύβˆš  π‘Žπ‘›π‘‘  π‘‘π‘Žπ‘›(𝐴–𝐡)=13β€Ύβˆš

; 0Β° < A + B ≀ 90Β°; A > B, find A and B.

π‘‘π‘Žπ‘›(𝐴+𝐡)⇒𝐴+π΅π΄π‘™π‘ π‘œ  π‘‘π‘Žπ‘›(𝐴–𝐡)β‡’π΄βˆ’π΅Adding (1) and (2)2𝐴⇒𝐴=3β€Ύβˆš=60  β€¦(1)=13β€Ύβˆš=30  β€¦(2)=90=45

Put the value of A in (1)

𝐴+𝐡⇒45+𝐡⇒𝐡=60=60=15

Answer: A = 45Β° and B = 15Β°

Click the following video to watch it’s solution.

Question 4

State whether the following are true or false. Justify your answer.

(i) sin (A + B) = sin A + sin B.Β 

Answer: False

Consider A = 30Β° and B = 60Β°

β‡’  π‘Žπ‘›π‘‘  π‘ π‘–𝑛𝐴+𝑠𝑖𝑛𝐡=𝑠𝑖𝑛30Β°+𝑠𝑖𝑛60Β°=12+3β€Ύβˆš2=1+3β€Ύβˆš2  β€¦  β€¦(1)𝑠𝑖𝑛(𝐴+𝐡)=𝑠𝑖𝑛(30Β°+60Β°)=𝑠𝑖𝑛90Β°=1  β€¦  β€¦  β€¦(2)

From (1) and (2)

𝑠𝑖𝑛𝐴+𝑠𝑖𝑛𝐡≠𝑠𝑖𝑛(𝐴+𝐡)

Hence the statement of not true in general.

(ii) The value of sin ΞΈ increases as ΞΈ increases.

Answer: Actually it’s answer is False but it is given as True in your book.

The reason is, the value of angle ΞΈ can be more than 90 degrees as well. The value of sin ΞΈ increases as ΞΈ increases from 0 to 90 degrees but sin ΞΈ starts decreasing as the value of ΞΈ increases from 90 to 180 degrees.Β 

Now talk to your class teacher to know what to write in exam πŸ˜‰

Ideally the question should be as follows:Β 

The value of sin ΞΈ increases as ΞΈ increases from 0 to 90 degrees.

(iii) The value of cos ΞΈ increases as ΞΈ increases.

Answer: False

Reason: cos 0Β° = 1 and cos 90Β° = 0

(iv) sin ΞΈ = cos ΞΈ for all values of ΞΈ.

Answer: False

Reason: sin 0Β° = 0Β and cos 0Β° = 1

(v) cot A is not defined for A = 0Β°.

Answer: True

Reason: Just remember the value of cot 0Β° or try to evaluate cos 0Β°/sin 0Β° πŸ˜‰

Click the following video to watch it’s solution.

Some general tips for solving trigonometry questions easily and quickly are:

  1. Prepare a table of all standard values of sin, cos, tan, cosec, sec and cot. This will help you memorize them to be applied quickly to your questions.
  2. Another important tip is that if you get stuck somewhere, try to change everything into sin and cos. In some questions it may help you progress further towards the solution.Β 
  3. Some important relations are sin(x) = cos(90-x) , tan(x) = cot(90-x) , cosec(x) = sec(90-x) and vice versa.
  4. It is important to know which side of the triangle is base and which is perpendicular. The one opposite to the angle under consideration is known as perpendicular and the one adjacent is the base. (the longest one is the hypotenuse)
  5. In case you are not able to prove LHS equal to RHS, then progress both of them separately and bring them to a common point. This will be considered a valid solution to the question.Β 

The post Exercise 8.2 (NCERT) – Introduction to Trigonometry, Class 10 – All Solutions appeared first on Mathemafia.

]]> Exercise 8.1 (NCERT) – Introduction to Trigonometry, Class 10 – All Solutions https://mathemafia.com/class-10-trigonometry-exercise-8-1-all-solutions-ncert/ Thu, 06 Feb 2025 11:51:34 +0000 https://mathemafia.com/?p=9799 Complete solution of Exercise 8.1, Chapter 8, Introduction to Trigonometry of NCERT book of Mathematics for Class 10. The solutions are written in pure mathematical way with mobile-friendly view, fully responsive. The free YouTube video tutorial of every question is embedded after the written solution. This educational content has been created with utmost care to make it easily readable and hence understandable by the students of class 10, especially preparing for the final board exams.

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Exercise 8.1 (NCERT) - Trigonometry, Class 10

Chapter 8 Exercise 8.1 Class 10

Below are the Quick links for all questions of Exercise 8.1, Trigonometry, Class 10.

Click a link to view the solution of corresponding question.

Below are the solutions of every question of exercise 8.1 of chapter 8, Introduction to Trigonometry, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.Β Β 

Question 1

In β–³ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :Β (i) sin A, cos A (ii) sin C, cos C

Question 1, Exercise 8.1, Chapter 8, Trigonometry
𝐴𝐡=24π‘π‘š,𝐡𝐢=7π‘π‘š

∡ β–³ ABC is a right angled triangle, let us apply pythagoras theorem

𝐴𝐢2𝐴𝐢=𝐴𝐡2+𝐡𝐢2=242+72=576+49=625=625β€Ύβ€Ύβ€Ύβ€Ύβˆš=25π‘π‘š

Β (i) Let us now find the trigonometric ratios sin A and cos A

When we take A as the angle, the perpendicular will be side opposite to it. So

BC – Perpendicular, AB – Base, AC – Hypotenuse

𝑠𝑖𝑛𝐴=π‘ƒπ‘’π‘Ÿπ‘π»π‘¦π‘=𝐡𝐢𝐴𝐢=725
π‘π‘œπ‘ π΄=π΅π‘Žπ‘ π‘’π»π‘¦π‘=𝐴𝐡𝐴𝐢=2425

(ii) Now we have to find the trigonometric ratios sin CΒ and cos C

When we take C as the angle, the perpendicular will be side opposite to it. So

AB – Perpendicular, BC – Base, AC – Hypotenuse

𝑠𝑖𝑛𝐢=π‘ƒπ‘’π‘Ÿπ‘π»π‘¦π‘=𝐴𝐡𝐴𝐢=2425
π‘π‘œπ‘ πΆ=π΅π‘Žπ‘ π‘’π»π‘¦π‘=𝐡𝐢𝐴𝐢=725
Click the following video to watch it’s solution.

Question 2

In Fig. 8.13, find tan P – cot R.

Question 2, Exercise 8.1, Chapter 8, Trigonometry
𝑃𝑄=12π‘π‘š,𝑃𝑅=13π‘π‘š

∡ β–³ PQR is a right angled triangle, let us apply pythagoras theorem and find QR.

𝑃𝑅2𝑄𝑅2𝑄𝑅=𝑃𝑄2+𝑄𝑅2=𝑃𝑅2–𝑃𝑄2=132–122=169–144=25=25β€Ύβ€Ύβ€Ύβˆš=5π‘π‘š

Let us now find the trigonometric ratios tan P and cot R

When we take P as the angle, the perpendicular will be side opposite to it. So

QR – Perpendicular, PQ – Base, PR – Hypotenuse

π‘‘π‘Žπ‘›π‘ƒ=π‘ƒπ‘’π‘Ÿπ‘π΅π‘Žπ‘ π‘’=𝑄𝑅𝑃𝑄=512…(1)

When we take R as the angle, the perpendicular will be side opposite to it. So

PQ – Perpendicular, QR – Base, PR – Hypotenuse

π‘π‘œπ‘‘π‘…=π΅π‘Žπ‘ π‘’π‘ƒπ‘’π‘Ÿπ‘=𝑄𝑅𝑃𝑄=512…(2)

From equations (1) and (2)

π‘‘π‘Žπ‘›π‘ƒβ€“π‘π‘œπ‘‘π‘…=512–512=0
Click the following video to watch it’s solution.

Question 3

If sinA = 3/4 calculate cos A and tan A.

Question 3, Exercise 8.1, Chapter 8, Trigonometry
𝑠𝑖𝑛𝐴=34

Now we know that

𝑠𝑖𝑛𝐴=π‘ƒπ‘’π‘Ÿπ‘π»π‘¦π‘=𝐡𝐢𝐴𝐢=34

Note: Now this is an important point to understand. Often students take Perpendicular (BC) as 3 and Hypotenuse (AC) as 4. But 3/4 is ratio of Perpendicular and Hypotenuse. Hence using the above ratio, we must solve it as follows.

𝐡𝐢𝐴𝐢=3π‘˜=4π‘˜

Now, let us apply Pythagoras theorem to find the third side (AB) of the right angled triangle.

𝐴𝐢2𝐴𝐡2𝐴𝐡=𝐴𝐡2+𝐡𝐢2=𝐴𝐢2–𝐡𝐢2=(4π‘˜)2–(3π‘˜)2=16π‘˜2–9π‘˜2=5π‘˜2=(5π‘˜2)β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=5β€Ύβˆšπ‘˜

Let us now calculate cos A and tan A

When we take A as the angle, the perpendicular will be the side opposite to it. Hence

BC – Perpendicular, AB – Base, AC – Hypotenuse

π‘π‘œπ‘ π΄=π΅π‘Žπ‘ π‘’π»π‘¦π‘=𝐴𝐡𝐴𝐢=5β€Ύβˆšπ‘˜4π‘˜=5β€Ύβˆš4
π‘‘π‘Žπ‘›π΄=π‘ƒπ‘’π‘Ÿπ‘π΅π‘Žπ‘ π‘’=𝐡𝐢𝐴𝐡=3π‘˜5β€Ύβˆšπ‘˜=35β€Ύβˆš
Click the following video to watch it’s solution.

Question 4

Given 15 cot A = 8, find sin A and sec A.

Question 4, Exercise 8.1, Chapter 8, Trigonometry

This question is similar to the previous question that we just solved.Β 

15Γ—π‘π‘œπ‘‘π΄π‘π‘œπ‘‘π΄=8=815

When we take A as the angle, the perpendicular will be side opposite to it. So

BC – Perpendicular, AB – Base, AC – Hypotenuse

Now we know that

π‘π‘œπ‘‘π΄=π΅π‘Žπ‘ π‘’π‘ƒπ‘’π‘Ÿπ‘=𝐴𝐡𝐡𝐢=815

Note: Now this is an important point to understand. Often students take Base (AB) as 8 and Perpendicular (BC) as 15. But 8/15 is ratio of Base and Perpendicular. Hence using the above ratio, we must solve it as follows.

⇒𝐴𝐡𝐡𝐢=8π‘˜=15π‘˜

Now, let us apply Pythagoras theorem to find the third side (AC) of the right angled triangle.

𝐴𝐢2=𝐴𝐡2+𝐡𝐢2=(8π‘˜)2+(15π‘˜)2=64π‘˜2+225π‘˜2=289π‘˜2=289π‘˜2β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=17π‘˜

Let us now calculate sin A and sec A

𝑠𝑖𝑛𝐴=π‘ƒπ‘’π‘Ÿπ‘π»π‘¦π‘=𝐡𝐢𝐴𝐢=15π‘˜17π‘˜=1517
𝑠𝑒𝑐𝐴=π»π‘¦π‘π΅π‘Žπ‘ π‘’=𝐴𝐢𝐴𝐡=17π‘˜8π‘˜=178
Click the following video to watch it’s solution.

Question 5

Given sec ΞΈ = 13/12 calculate all other trigonometric ratios.

Super Tip: We have to find all other trigonometry ratios in this question and for that we need to know the dimensions of all 3 sides of the triangle. But before that, we can directly find the value of cos θ. This way, you will quickly score 20% marks 😊

Question 5, Exercise 8.1, Chapter 8, Trigonometry

cos ΞΈ

π‘π‘œπ‘ ΞΈ=1𝑠𝑒𝑐θ=11312=1213

Let us consider that angle ΞΈ is at vertex A of right angled triangle ABC with right angle at B. This means

BC – Perpendicular, AB – Base, AC – HypotenuseΒ 

𝑠𝑒𝑐θ=π»π‘¦π‘π΅π‘Žπ‘ π‘’=𝐴𝐢𝐴𝐡=1312

Note: Now this is an important point to understand. Often students take Hypotenuse (AC) as 13 and Base (AB) as 12. But 13/12 is ratio of Hypotenuse and Base. Hence using the above ratio, we must solve it as follows.

⇒𝐴𝐢𝐴𝐡=13π‘˜=12π‘˜

Now, let us apply Pythagoras theorem to find the third side (BC) of the right angled triangle.

𝐴𝐢2𝐡𝐢2=𝐴𝐡2+𝐡𝐢2=𝐴𝐢2–𝐴𝐡2=(13π‘˜)2–(12π‘˜)2=169π‘˜2–144π‘˜2=49π‘˜2=(49π‘˜2)β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=7π‘˜

sin ΞΈ

𝑠𝑖𝑛θ=π‘ƒπ‘’π‘Ÿπ‘π»π‘¦π‘=𝐡𝐢𝐴𝐢=5π‘˜13π‘˜=713

cosec ΞΈ

Now cosec ΞΈ can be calculated in 2 ways. (Write only 1 of these methods in exam)

Method 1

π‘π‘œπ‘ π‘’π‘ΞΈ=π»π‘¦π‘π‘ƒπ‘’π‘Ÿπ‘=𝐴𝐢𝐡𝐢=13π‘˜7π‘˜=137

Method 2

π‘π‘œπ‘ π‘’π‘ΞΈ=1𝑠𝑖𝑛θ=1713=137

tan ΞΈ

Now tan ΞΈ can be calculated in 2 ways. (WriteΒ only 1 of these methods in exam)

Method 1

π‘‘π‘Žπ‘›ΞΈ=π‘ƒπ‘’π‘Ÿπ‘π΅π‘Žπ‘ π‘’=𝐡𝐢𝐴𝐡=7π‘˜12π‘˜=712

Method 2

π‘‘π‘Žπ‘›ΞΈ=π‘ π‘–π‘›ΞΈπ‘π‘œπ‘ ΞΈ=7131213=712

cot ΞΈ

Now cot ΞΈ can be calculated in 2 ways. (WriteΒ only 1 of these methods in exam)

Method 1

π‘π‘œπ‘‘ΞΈ=π΅π‘Žπ‘ π‘’π‘ƒπ‘’π‘Ÿπ‘=𝐴𝐡𝐡𝐢=12π‘˜7π‘˜=127

Method 2

π‘π‘œπ‘‘ΞΈ=π‘π‘œπ‘ ΞΈπ‘ π‘–π‘›ΞΈ=1213713=127
Click the following video to watch it’s solution.

Question 6

If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Copyright - Mathemafia Question 6, Exercise 8.1, Chapter 8, Trigonometry

It is given that ∠A and ∠B are acute angles

β‡’βˆ πΆ=90Β°

Now for angle A

PerpendicularΒ = BC (side opp. to vertex A)

Base = AC (side adjacent to vertex A)

& for angle B

PerpendicularΒ = AC (side opp. to vertex B)

Base = BC (side opposite to vertex B)

Hypotenuse (AB) will remain same as hypotenuse in a right angled triangle is the longest side.Β 

HenceΒ 

π‘π‘œπ‘ π΄π‘π‘œπ‘ π΅=𝐴𝐢𝐴𝐡…(1)=𝐡𝐢𝐴𝐡…(2)

But we know thatΒ 

cosA = cosB

From (1) and (2)

𝐴𝐢𝐴𝐡𝐴𝐢=𝐡𝐢𝐴𝐡=𝐡𝐢

This implies β–³ABC is an isosceles triangle.Β 

β‡’βˆ π΄=∠𝐡
Click the following video to watch it’s solution.

Question 7

If cot ΞΈ = 7/8 evaluate :

(𝑖)  (1+𝑠𝑖𝑛θ)(1–𝑠𝑖𝑛θ)(1+π‘π‘œπ‘ ΞΈ)(1β€“π‘π‘œπ‘ ΞΈ)(𝑖𝑖)  π‘π‘œπ‘‘2ΞΈ

Time saver tip:Β Such questions are very easy to solve. No need to find any other trigonometric ratio to solve this question.

(𝑖𝑖)  π‘π‘œπ‘‘2ΞΈ
π‘π‘œπ‘‘2ΞΈ  π‘Žπ‘π‘‘π‘’π‘Žπ‘™π‘™π‘¦  π‘–𝑠  π‘π‘œπ‘‘ΞΈΓ—π‘π‘œπ‘‘ΞΈ
=(π‘π‘œπ‘‘ΞΈ)2=(78)2=4964
(𝑖)  (1+𝑠𝑖𝑛θ)(1–𝑠𝑖𝑛θ)(1+π‘π‘œπ‘ ΞΈ)(1β€“π‘π‘œπ‘ ΞΈ)
=1–𝑠𝑖𝑛2ΞΈ1β€“π‘π‘œπ‘ 2ΞΈ=π‘π‘œπ‘ 2θ𝑠𝑖𝑛2ΞΈ=π‘π‘œπ‘‘2ΞΈ
=(78)2=4964
Click the following video to watch it’s solution.

Question 8

If 3 cot A = 4, check whether

1β€“π‘‘π‘Žπ‘›2𝐴1+π‘‘π‘Žπ‘›2𝐴=π‘π‘œπ‘ 2𝐴–𝑠𝑖𝑛2𝐴  π‘œπ‘Ÿ  π‘›π‘œπ‘‘.
Question 8, Exercise 8.1, Chapter 8, Trigonometry
3Γ—π‘π‘œπ‘‘π΄π‘π‘œπ‘‘π΄=4=43

When we take A as the angle, then

BC – Perpendicular, AB – Base, AC – Hypotenuse

Now we know that

π‘π‘œπ‘‘π΄=π΅π‘Žπ‘ π‘’π‘ƒπ‘’π‘Ÿπ‘=𝐴𝐡𝐡𝐢=43

Note: Now this is an important point to understand. Often students take Base (AB) as 4 and Perpendicular (BC) as 3. But 4/3 is ratio of Base and Perpendicular. Hence using the above ratio, we must solve it as follows.

⇒𝐴𝐡𝐡𝐢=4π‘˜=3π‘˜

Applying Pythagoras theorem

𝐴𝐢2𝐴𝐢=𝐴𝐡2+𝐡𝐢2=(4π‘˜)2+(3π‘˜)2=16π‘˜2+9π‘˜2=25π‘˜2=(25π‘˜2)β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=5π‘˜

Let’s calculate the values of sin A, cos A & tan A

𝑠𝑖𝑛𝐴=π‘ƒπ‘’π‘Ÿπ‘π»π‘¦π‘=𝐡𝐢𝐴𝐢=3π‘˜5π‘˜=35π‘π‘œπ‘ π΄=π΅π‘Žπ‘ π‘’π»π‘¦π‘=𝐴𝐡𝐴𝐢=4π‘˜5π‘˜=45π‘‘π‘Žπ‘›π΄=1π‘π‘œπ‘‘π΄=14/3=34
𝐿𝐻𝑆=1β€“π‘‘π‘Žπ‘›2𝐴1+π‘‘π‘Žπ‘›2𝐴=1–(3/4)21+(3/4)2=1–9/161+9/16=7/1625/16=725
𝑅𝐻𝑆=π‘π‘œπ‘ 2𝐴–𝑠𝑖𝑛2𝐴=(4/5)2–(3/5)2=16/25–9/25=725=𝐿𝐻𝑆

Hence proved.

Click the following video to watch it’s solution.

Question 9

In triangle ABC, right-angled at B, if tan A = 1/√3 find the value of:

(i) sin A cos C + cos A sin C

(ii) cos A cos C – sin A sin C

Question 9, Exercise 8.1, Chapter 8, Trigonometry
π‘‘π‘Žπ‘›π΄=13β€Ύβˆš

When we take A as the angle, the perpendicular will be side opposite to it. So

BC – Perpendicular, AB – Base, AC – Hypotenuse

Now we know that

π‘‘π‘Žπ‘›π΄=π‘ƒπ‘’π‘Ÿπ‘π΅π‘Žπ‘ π‘’=𝐡𝐢𝐴𝐡=13β€Ύβˆšπ»π‘’π‘›π‘π‘’  π΅πΆ=π‘˜,𝐴𝐡=3β€Ύβˆšπ‘˜

If you have any doubt about the above statement, read the first NOTE written on this page.

Now, using Pythagoras theorem

𝐴𝐢2𝐴𝐢=𝐴𝐡2+𝐡𝐢2=π‘˜2+(3β€Ύβˆšπ‘˜)2=π‘˜2+3π‘˜2=4π‘˜2=(4π‘˜2)β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš=2π‘˜

(i) sin A cos C + cos A sin C

𝑠𝑖𝑛𝐴=𝐡𝐢𝐴𝐢=π‘˜2π‘˜=12π‘π‘œπ‘ π΄=𝐴𝐡𝐴𝐢=3β€Ύβˆšπ‘˜2π‘˜=3β€Ύβˆš2

When we take C as the angle, the perpendicular will be side opposite to it. So

AB – Perpendicular, BC – Base, AC – Hypotenuse

𝑠𝑖𝑛𝐢=𝐴𝐡𝐴𝐢=3β€Ύβˆšπ‘˜2π‘˜=3β€Ύβˆš2π‘π‘œπ‘ πΆ=𝐡𝐢𝐴𝐢=π‘˜2π‘˜=12π‘ π‘–π‘›π΄π‘π‘œπ‘ πΆ+π‘π‘œπ‘ π΄π‘ π‘–π‘›πΆ
=12Γ—12+3β€Ύβˆš2Γ—3β€Ύβˆš2=14+34=1

(ii) cos A cos C – sin A sin C

π‘π‘œπ‘ π΄π‘π‘œπ‘ πΆβ€“π‘ π‘–π‘›π΄π‘ π‘–π‘›πΆ
=3β€Ύβˆš2Γ—12–12Γ—3β€Ύβˆš2=3β€Ύβˆš4–3β€Ύβˆš4=0
Click the following video to watch it’s solution.

Question 10

In β–³PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Question 10, Exercise 8.1, Chapter 8, Trigonometry
𝐿𝑒𝑑  π‘„𝑅𝑃𝑅+𝑄𝑅⇒𝑃𝑅=π‘₯=25=25–𝑄𝑅=25βˆ’π‘₯

Applying Pythagoras theorem

𝑃𝑅2(25–π‘₯)2β€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβ€Ύβˆš625+π‘₯2–50π‘₯625–50π‘₯50π‘₯π‘₯=𝑄𝑅2+𝑃𝑄2=(5)2+(π‘₯)2=25+π‘₯2=25=625–25=600=600/50=12
⇒𝑄𝑅𝑃𝑅=π‘₯=12π‘π‘š=25–𝑄𝑅=25–12=13π‘π‘š

Let’s calculate the values of sin P, cos P & tan P

𝑠𝑖𝑛𝑃=π‘ƒπ‘’π‘Ÿπ‘π»π‘¦π‘=𝑄𝑅𝑃𝑅=1213π‘π‘œπ‘ π‘ƒ=π΅π‘Žπ‘ π‘’π»π‘¦π‘=𝑃𝑄𝑃𝑅=513π‘‘π‘Žπ‘›π‘ƒ=π‘ƒπ‘’π‘Ÿπ‘π΅π‘Žπ‘ π‘’=𝑄𝑅𝑃𝑄=125
Click the following video to watch it’s solution.

Question 11

State whether the following are true or false. Justify your answer.

(i) The value of tan A is always less than 1.

(ii) sec A = 12/5 for some value of angle A.

(iii) cos A is the abbreviation used for the cosecant of angle A.

(iv) cot A is the product of cot and A.

(v) sin ΞΈ = 4/3 for some angle ΞΈ

Solution:

(i) The value of tan A is always less than 1 – False

Reason: We know that tan 60° = √3 and √3 > 1, hence this statement is false.

(i) sec A = 12/5 for some value of angle A. – True

Reason: Trigonometric ratio sec is ratio of Hyp/Base and we know that Hypotenuse is always the longest side of a right angled triangle. Hence taking Hyp = 12 and Base = 5 we can get a right angled triangle.Β 

(iii) cos A is the abbreviation used for the cosecant of angle A – False

Reason: cos A is abbreviation of cosine of angle A.

(iv) cot A is the product of cot and A – False

Reason: cot A is single a term and it is cotangent of A.

(v) sin ΞΈ = 4/3 for some angle ΞΈ – False

Reason: Value of sin ΞΈ can never be greater than 1 because it is the ratio of Perpendicular and Hypotenuse and perpendicular can never be longer than hypotenuse.

Click the following video to watch it’s solution.
\(\begin{aligned}Β  \end{aligned}\)

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