exercise 8.3 solved Archives – Mathemafia https://mathemafia.com/tag/exercise-8-3-solved/ Mathematics is like music - Enjoy it!!! Sat, 08 Mar 2025 07:05:20 +0000 en-US hourly 1 https://wordpress.org/?v=7.0.1 https://mathemafia.com/wp-content/uploads/2020/08/cropped-M__Logo__SM-1-32x32.png exercise 8.3 solved Archives – Mathemafia https://mathemafia.com/tag/exercise-8-3-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.

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

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