Exercise 8.3 (NCERT) – Introduction to Trigonometry, Class 10 – All Solutions

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
• Ex: 8.3, Question 1
• Ex: 8.3, Question 2
• Ex: 8.3 Question 3
  • Part (i)
  • Part (ii)
  • Part (iii)
  • Part (iv)

• Ex: 8.3, Question 4
  • Part (i)
  • Part (ii)
  • Part (iii)
  • Part (iv)
  • Part (v)
  • Part (vi)
  • Part (vii)
  • Part (viii)
  • Part (ix)
  • Part (x)
• Recommended Tutorial

3 Idiots 😜 Identities – Don’t forget

(1)  𝑠𝑖𝑛2𝐴+𝑐𝑜𝑠2𝐴=1(2)  1+𝑡𝑎𝑛2𝐴=𝑠𝑒𝑐2𝐴(3)  1+𝑐𝑜𝑡2𝐴=𝑐𝑜𝑠𝑒𝑐2𝐴

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. 

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

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. 

𝑡𝑎𝑛𝐴=1𝑐𝑜𝑡𝐴

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+𝑡𝑎𝑛2𝐴=1+1𝑐𝑜𝑡2𝐴=1+𝑐𝑜𝑡2𝐴𝑐𝑜𝑡2𝐴=1+𝑐𝑜𝑡2𝐴𝑐𝑜𝑡2𝐴‾‾‾‾‾‾‾‾‾‾√

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 that 

𝑐𝑜𝑠𝐴𝑐𝑜𝑠𝐴=1𝑠𝑒𝑐𝐴=𝑐𝑜𝑡2𝐴1+𝑐𝑜𝑡2𝐴‾‾‾‾‾‾‾‾‾‾√

Use the value of cos A in the following identity

𝑠𝑖𝑛2𝐴+𝑐𝑜𝑠2𝐴=1

⇒𝑠𝑖𝑛2𝐴=1−𝑐𝑜𝑠2𝐴=1–(𝑐𝑜𝑡2𝐴1+𝑐𝑜𝑡2𝐴‾‾‾‾‾‾‾‾‾‾√)2=1–𝑐𝑜𝑡2𝐴1+𝑐𝑜𝑡2𝐴=1+𝑐𝑜𝑡2𝐴–𝑐𝑜𝑡2𝐴1+𝑐𝑜𝑡2𝐴=11+𝑐𝑜𝑡2𝐴

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. 

𝑐𝑜𝑠𝐴=1𝑠𝑒𝑐𝐴

sin A

We know that

⇒  ⇒  𝑠𝑖𝑛2𝐴+𝑐𝑜𝑠2𝐴=1𝑠𝑖𝑛2𝐴=1–𝑐𝑜𝑠2𝐴=1–1𝑠𝑒𝑐2𝐴=𝑠𝑒𝑐2𝐴−1𝑠𝑒𝑐2𝐴𝑠𝑖𝑛𝐴=𝑠𝑒𝑐2𝐴−1𝑠𝑒𝑐2𝐴‾‾‾‾‾‾‾‾‾‾√=𝑠𝑒𝑐2𝐴−1‾‾‾‾‾‾‾‾‾√𝑠𝑒𝑐𝐴

cosec A

Now cosec A is reciprocal of sin A, hence

𝑐𝑜𝑠𝑒𝑐𝐴=1𝑠𝑖𝑛𝐴=𝑠𝑒𝑐𝐴𝑠𝑒𝑐2𝐴−1‾‾‾‾‾‾‾‾‾√

tan A

Method 1

Use Identity1+𝑡𝑎𝑛2𝐴=𝑠𝑒𝑐2𝐴

⇒  ⇒  𝑡𝑎𝑛2𝐴=𝑠𝑒𝑐2𝐴–1𝑡𝑎𝑛𝐴=𝑠𝑒𝑐2𝐴–1‾‾‾‾‾‾‾‾√

Method 2

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

𝑡𝑎𝑛𝐴=𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴=𝑠𝑒𝑐2𝐴−1√𝑠𝑒𝑐𝐴1𝑠𝑒𝑐𝐴=𝑠𝑒𝑐2𝐴−1‾‾‾‾‾‾‾‾‾√𝑠𝑒𝑐𝐴×𝑠𝑒𝑐𝐴=𝑠𝑒𝑐2𝐴−1‾‾‾‾‾‾‾‾‾√

cot A

Now cot A is reciprocal of tan A, hence

𝑐𝑜𝑡𝐴=1𝑡𝑎𝑛𝐴=1𝑠𝑒𝑐2𝐴–1‾‾‾‾‾‾‾‾√

Question 3

Choose the correct option. Justify your choice.

(𝑖)  9  𝑠𝑒𝑐2𝐴  –  9  𝑡𝑎𝑛2𝐴=(𝐴)  1  (𝐵)  9  (𝐶)  8  (𝐷)  0

=  9  𝑠𝑒𝑐2𝐴  –  9  𝑡𝑎𝑛2𝐴  9×(𝑠𝑒𝑐2𝐴  –  𝑡𝑎𝑛2𝐴)

⇒  We know1+𝑡𝑎𝑛2𝐴=𝑠𝑒𝑐2𝐴𝑠𝑒𝑐2𝐴–𝑡𝑎𝑛2𝐴=1

==  9×1  9

Answer: (B)

(𝑖𝑖)  (1+𝑡𝑎𝑛θ+𝑠𝑒𝑐θ)(1+𝑐𝑜𝑡θ–𝑐𝑜𝑠𝑒𝑐θ)=(𝐴)  0  (𝐵)  1  (𝐶)  2  (𝐷)  –1

(1+𝑡𝑎𝑛θ+𝑠𝑒𝑐θ)(1+𝑐𝑜𝑡θ–𝑐𝑜𝑠𝑒𝑐θ)=(1+𝑠𝑖𝑛θ𝑐𝑜𝑠θ+1𝑐𝑜𝑠θ)(1+𝑐𝑜𝑠θ𝑠𝑖𝑛θ–1𝑠𝑖𝑛θ)=(𝑐𝑜𝑠θ+𝑠𝑖𝑛θ+1𝑐𝑜𝑠θ)(𝑠𝑖𝑛θ+𝑐𝑜𝑠θ–1𝑠𝑖𝑛θ)=(𝑠𝑖𝑛θ+𝑐𝑜𝑠θ+1𝑐𝑜𝑠θ)(𝑠𝑖𝑛θ+𝑐𝑜𝑠θ–1𝑠𝑖𝑛θ)=(𝑠𝑖𝑛θ+𝑐𝑜𝑠θ)2–1𝑠𝑖𝑛θ  𝑐𝑜𝑠θ=𝑠𝑖𝑛2θ+𝑐𝑜𝑠2θ+2 sinθ  𝑐𝑜𝑠θ–1𝑠𝑖𝑛θ  𝑐𝑜𝑠θ=1+2 sinθ  𝑐𝑜𝑠θ–1𝑠𝑖𝑛θ  𝑐𝑜𝑠θ=2 sinθ  𝑐𝑜𝑠θ𝑠𝑖𝑛θ  𝑐𝑜𝑠θ=2

Answer: (C)

(𝑖𝑖𝑖)  (𝑠𝑒𝑐𝐴+𝑡𝑎𝑛𝐴)(1–𝑠𝑖𝑛𝐴)=(𝐴)  𝑠𝑒𝑐𝐴  (𝐵)  𝑠𝑖𝑛𝐴(𝐶)  𝑐𝑜𝑠𝑒𝑐𝐴  (𝐷)  𝑐𝑜𝑠𝐴

Let us convert all terms to sin and cos

(𝑠𝑒𝑐𝐴+𝑡𝑎𝑛𝐴)(1  –  𝑠𝑖𝑛𝐴)=(1𝑐𝑜𝑠𝐴+𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴)(1  –  𝑠𝑖𝑛𝐴)=(1+𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴)(1  –  𝑠𝑖𝑛𝐴)=1–𝑠𝑖𝑛2𝐴𝑐𝑜𝑠𝐴=𝑐𝑜𝑠2𝐴𝑐𝑜𝑠𝐴=𝑐𝑜𝑠𝐴

Answer: (D)

(𝑖𝑣)  1+𝑡𝑎𝑛2𝐴1+𝑐𝑜𝑡2𝐴=(𝐴)  𝑠𝑒𝑐2𝐴  (𝐵)  −1(𝐶)  𝑐𝑜𝑡2𝐴  (𝐷) tan2𝐴

Method 1:

1+𝑡𝑎𝑛2𝐴1+𝑐𝑜𝑡2𝐴=1+𝑡𝑎𝑛2𝐴1+1𝑡𝑎𝑛2𝐴=1+𝑡𝑎𝑛2𝐴𝑡𝑎𝑛2𝐴+1𝑡𝑎𝑛2𝐴=(1+𝑡𝑎𝑛2𝐴)×𝑡𝑎𝑛2𝐴(1+𝑡𝑎𝑛2𝐴)=𝑡𝑎𝑛2𝐴

Method 2:

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

1+𝑡𝑎𝑛2𝐴1+𝑐𝑜𝑡2𝐴=1+𝑠𝑖𝑛2𝐴𝑐𝑜𝑠2𝐴1+𝑐𝑜𝑠2𝐴𝑠𝑖𝑛2𝐴=𝑐𝑜𝑠2𝐴+𝑠𝑖𝑛2𝐴𝑐𝑜𝑠2𝐴𝑠𝑖𝑛2𝐴+𝑐𝑜𝑠2𝐴𝑠𝑖𝑛2𝐴=1𝑐𝑜𝑠2𝐴1𝑠𝑖𝑛2𝐴=1𝑐𝑜𝑠2𝐴×𝑠𝑖𝑛2𝐴=𝑡𝑎𝑛2𝐴

Answer: (D)

Question 4

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

(𝑖)  (𝑐𝑜𝑠𝑒𝑐θ–𝑐𝑜𝑡θ)2=1−𝑐𝑜𝑠θ1+𝑐𝑜𝑠θ

𝐿𝐻𝑆=(𝑐𝑜𝑠𝑒𝑐θ–𝑐𝑜𝑡θ)2=(1𝑠𝑖𝑛θ–𝑐𝑜𝑠θ𝑠𝑖𝑛θ)2=(1−𝑐𝑜𝑠θ𝑠𝑖𝑛θ)2=(1−𝑐𝑜𝑠θ)2𝑠𝑖𝑛2θ=(1−𝑐𝑜𝑠θ)21−𝑐𝑜𝑠2θ=(1−𝑐𝑜𝑠θ)2(1−𝑐𝑜𝑠θ)(1+𝑐𝑜𝑠θ)=1−𝑐𝑜𝑠θ1+𝑐𝑜𝑠θ=𝑅𝐻𝑆

(𝑖𝑖)  𝑐𝑜𝑠𝐴1+𝑠𝑖𝑛𝐴+1+𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴=2  𝑠𝑒𝑐𝐴

𝐿𝐻𝑆=𝑐𝑜𝑠𝐴1+𝑠𝑖𝑛𝐴+1+𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴=𝑐𝑜𝑠2𝐴+(1+𝑠𝑖𝑛𝐴)2(1+𝑠𝑖𝑛𝐴)(𝑐𝑜𝑠𝐴)=𝑐𝑜𝑠2𝐴+1+𝑠𝑖𝑛2𝐴+2𝑠𝑖𝑛𝐴(1+𝑠𝑖𝑛𝐴)(𝑐𝑜𝑠𝐴)=𝑠𝑖𝑛2𝐴+𝑐𝑜𝑠2𝐴+1+2𝑠𝑖𝑛𝐴(1+𝑠𝑖𝑛𝐴)(𝑐𝑜𝑠𝐴)=1+1+2𝑠𝑖𝑛𝐴(1+𝑠𝑖𝑛𝐴)(𝑐𝑜𝑠𝐴)=2+2𝑠𝑖𝑛𝐴(1+𝑠𝑖𝑛𝐴)(𝑐𝑜𝑠𝐴)=2(1+𝑠𝑖𝑛𝐴)(1+𝑠𝑖𝑛𝐴)(𝑐𝑜𝑠𝐴)=2𝑐𝑜𝑠𝐴=2𝑠𝑒𝑐𝐴=𝑅𝐻𝑆

(𝑖𝑖𝑖)  𝑡𝑎𝑛θ1−𝑐𝑜𝑡θ+𝑐𝑜𝑡θ1−𝑡𝑎𝑛θ=1+𝑠𝑒𝑐θ  𝑐𝑜𝑠𝑒𝑐θ

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

𝐿𝐻𝑆=𝑡𝑎𝑛θ1−𝑐𝑜𝑡θ+𝑐𝑜𝑡θ1−𝑡𝑎𝑛θ=𝑠𝑖𝑛θ𝑐𝑜𝑠θ1−𝑐𝑜𝑠θ𝑠𝑖𝑛θ+𝑐𝑜𝑠θ𝑠𝑖𝑛θ1−𝑠𝑖𝑛θ𝑐𝑜𝑠θ=𝑠𝑖𝑛θ𝑐𝑜𝑠θ𝑠𝑖𝑛θ–𝑐𝑜𝑠θ𝑠𝑖𝑛θ+𝑐𝑜𝑠θ𝑠𝑖𝑛θ𝑐𝑜𝑠θ−𝑠𝑖𝑛θ𝑐𝑜𝑠θ=𝑠𝑖𝑛θ𝑐𝑜𝑠θ×𝑠𝑖𝑛θ𝑠𝑖𝑛θ–𝑐𝑜𝑠θ+𝑐𝑜𝑠θ𝑠𝑖𝑛θ×𝑐𝑜𝑠θ𝑐𝑜𝑠θ−𝑠𝑖𝑛θ=𝑠𝑖𝑛2θ𝑐𝑜𝑠θ×(𝑠𝑖𝑛θ–𝑐𝑜𝑠θ)+𝑐𝑜𝑠2θ𝑠𝑖𝑛θ×(𝑐𝑜𝑠θ−𝑠𝑖𝑛θ)=𝑠𝑖𝑛2θ𝑐𝑜𝑠θ×(𝑠𝑖𝑛θ–𝑐𝑜𝑠θ)–𝑐𝑜𝑠2θ𝑠𝑖𝑛θ×(𝑠𝑖𝑛θ−𝑐𝑜𝑠θ)=𝑠𝑖𝑛3θ–𝑐𝑜𝑠3θ𝑐𝑜𝑠θ×𝑠𝑖𝑛θ×(𝑠𝑖𝑛θ–𝑐𝑜𝑠θ)=(𝑠𝑖𝑛θ–𝑐𝑜𝑠θ)×(𝑠𝑖𝑛2θ+𝑐𝑜𝑠2θ+𝑠𝑖𝑛θ×𝑐𝑜𝑠θ)𝑐𝑜𝑠θ×𝑠𝑖𝑛θ×(𝑠𝑖𝑛θ–𝑐𝑜𝑠θ)=1+𝑠𝑖𝑛θ×𝑐𝑜𝑠θ𝑐𝑜𝑠θ×𝑠𝑖𝑛θ=1𝑐𝑜𝑠θ×𝑠𝑖𝑛θ+𝑠𝑖𝑛θ×𝑐𝑜𝑠θ𝑐𝑜𝑠θ×𝑠𝑖𝑛θ=𝑠𝑒𝑐θ×𝑐𝑜𝑠𝑒𝑐θ+1=1+𝑠𝑒𝑐θ×𝑐𝑜𝑠𝑒𝑐θ=𝑅𝐻𝑆

(𝑖𝑣)  1+𝑠𝑒𝑐𝐴𝑠𝑒𝑐𝐴=𝑠𝑖𝑛2𝐴1−𝑐𝑜𝑠𝐴

[Hint : Simplify LHS and RHS separately]

𝐿𝐻𝑆=1+𝑠𝑒𝑐𝐴𝑠𝑒𝑐𝐴=1+1𝑐𝑜𝑠𝐴1𝑐𝑜𝑠𝐴=𝑐𝑜𝑠𝐴+1𝑐𝑜𝑠𝐴1𝑐𝑜𝑠𝐴=𝑐𝑜𝑠𝐴+1𝑐𝑜𝑠𝐴×𝑐𝑜𝑠𝐴1=1+𝑐𝑜𝑠𝐴  …  …(1)

𝑅𝐻𝑆=𝑠𝑖𝑛2𝐴1−𝑐𝑜𝑠𝐴=1−𝑐𝑜𝑠2𝐴1−𝑐𝑜𝑠𝐴=(1−𝑐𝑜𝑠𝐴)(1+𝑐𝑜𝑠𝐴)1−𝑐𝑜𝑠𝐴=1+𝑐𝑜𝑠𝐴  …  …(2)

From (1) and (2)

LHS = RHS

(𝑣)  𝑐𝑜𝑠𝐴–𝑠𝑖𝑛𝐴+1𝑐𝑜𝑠𝐴+𝑠𝑖𝑛𝐴–1=𝑐𝑜𝑠𝑒𝑐𝐴+𝑐𝑜𝑡𝐴using the identity𝑐𝑜𝑠𝑒𝑐2𝐴=1+𝑐𝑜𝑡2𝐴

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

𝐿𝐻𝑆=𝑐𝑜𝑠𝐴–𝑠𝑖𝑛𝐴+1𝑐𝑜𝑠𝐴+𝑠𝑖𝑛𝐴–1

As we are asked to use the identity

𝑐𝑜𝑠𝑒𝑐2𝐴=1+𝑐𝑜𝑡2𝐴

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

𝐿𝐻𝑆=𝑐𝑜𝑠𝐴–𝑠𝑖𝑛𝐴+1𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴+𝑠𝑖𝑛𝐴–1𝑠𝑖𝑛𝐴=𝑐𝑜𝑡𝐴–1+𝑐𝑜𝑠𝑒𝑐𝐴𝑐𝑜𝑡𝐴+1–𝑐𝑜𝑠𝑒𝑐𝐴=𝑐𝑜𝑠𝑒𝑐𝐴+𝑐𝑜𝑡𝐴–1𝑐𝑜𝑡𝐴+1–𝑐𝑜𝑠𝑒𝑐𝐴  …(1)

From identity𝑐𝑜𝑠𝑒𝑐2𝐴=1+𝑐𝑜𝑡2𝐴We get1=𝑐𝑜𝑠𝑒𝑐2𝐴–𝑐𝑜𝑡2𝐴Let us use it (1)

𝐿𝐻𝑆=(𝑐𝑜𝑠𝑒𝑐𝐴+𝑐𝑜𝑡𝐴)–(𝑐𝑜𝑠𝑒𝑐2𝐴−𝑐𝑜𝑡2𝐴)𝑐𝑜𝑡𝐴+1–𝑐𝑜𝑠𝑒𝑐𝐴=(𝑐𝑜𝑠𝑒𝑐𝐴+𝑐𝑜𝑡𝐴)–(𝑐𝑜𝑠𝑒𝑐𝐴−𝑐𝑜𝑡𝐴)(𝑐𝑜𝑠𝑒𝑐𝐴+𝑐𝑜𝑡𝐴)1+𝑐𝑜𝑡𝐴–𝑐𝑜𝑠𝑒𝑐𝐴=(𝑐𝑜𝑠𝑒𝑐𝐴+𝑐𝑜𝑡𝐴)[1−(𝑐𝑜𝑠𝑒𝑐𝐴−𝑐𝑜𝑡𝐴)]1+𝑐𝑜𝑡𝐴–𝑐𝑜𝑠𝑒𝑐𝐴=(𝑐𝑜𝑠𝑒𝑐𝐴+𝑐𝑜𝑡𝐴)(1−𝑐𝑜𝑠𝑒𝑐𝐴+𝑐𝑜𝑡𝐴)1+𝑐𝑜𝑡𝐴–𝑐𝑜𝑠𝑒𝑐𝐴=(𝑐𝑜𝑠𝑒𝑐𝐴+𝑐𝑜𝑡𝐴)(1+𝑐𝑜𝑡𝐴−𝑐𝑜𝑠𝑒𝑐𝐴)1+𝑐𝑜𝑡𝐴–𝑐𝑜𝑠𝑒𝑐𝐴=𝑐𝑜𝑠𝑒𝑐𝐴+𝑐𝑜𝑡𝐴=𝑅𝐻𝑆

(𝑣𝑖)  1+𝑠𝑖𝑛𝐴1−𝑠𝑖𝑛𝐴‾‾‾‾‾‾‾‾‾√=𝑠𝑒𝑐𝐴+𝑡𝑎𝑛𝐴

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

𝐿𝐻𝑆=1+𝑠𝑖𝑛𝐴1−𝑠𝑖𝑛𝐴‾‾‾‾‾‾‾‾‾√=1+𝑠𝑖𝑛𝐴1−𝑠𝑖𝑛𝐴×1+𝑠𝑖𝑛𝐴1+𝑠𝑖𝑛𝐴‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√=(1+𝑠𝑖𝑛𝐴)21−𝑠𝑖𝑛2𝐴‾‾‾‾‾‾‾‾‾‾‾‾√=(1+𝑠𝑖𝑛𝐴)2𝑐𝑜𝑠2𝐴‾‾‾‾‾‾‾‾‾‾‾‾√=1+𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴=1𝑐𝑜𝑠𝐴+𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴=𝑠𝑒𝑐𝐴+𝑡𝑎𝑛𝐴=𝑅𝐻𝑆

(𝑣𝑖𝑖)  𝑠𝑖𝑛θ−2𝑠𝑖𝑛3θ2𝑐𝑜𝑠3θ−𝑐𝑜𝑠θ=𝑡𝑎𝑛θ

𝐿𝐻𝑆=𝑠𝑖𝑛θ−2  𝑠𝑖𝑛3θ2  𝑐𝑜𝑠3θ−𝑐𝑜𝑠θ=𝑠𝑖𝑛θ  (1−2  𝑠𝑖𝑛2θ)𝑐𝑜𝑠θ  (2  𝑐𝑜𝑠2θ−1)=𝑠𝑖𝑛θ  (𝑠𝑖𝑛2θ+𝑐𝑜𝑠2θ−2  𝑠𝑖𝑛2θ)𝑐𝑜𝑠θ  [2  𝑐𝑜𝑠2θ−(𝑠𝑖𝑛2θ+𝑐𝑜𝑠2θ)]=𝑠𝑖𝑛θ  (𝑐𝑜𝑠2θ−𝑠𝑖𝑛2θ)𝑐𝑜𝑠θ  (𝑐𝑜𝑠2θ−𝑠𝑖𝑛2θ)=𝑠𝑖𝑛θ𝑐𝑜𝑠θ=𝑡𝑎𝑛θ=𝑅𝐻𝑆

(𝑣𝑖𝑖𝑖)  (𝑠𝑖𝑛𝐴+𝑐𝑜𝑠𝑒𝑐𝐴)2+(𝑐𝑜𝑠𝐴+𝑠𝑒𝑐𝐴)2=7+𝑡𝑎𝑛2𝐴+𝑐𝑜𝑡2𝐴

𝐿𝐻𝑆=(𝑠𝑖𝑛𝐴+𝑐𝑜𝑠𝑒𝑐𝐴)2+(𝑐𝑜𝑠𝐴+𝑠𝑒𝑐𝐴)2=𝑠𝑖𝑛2𝐴+𝑐𝑜𝑠𝑒𝑐2𝐴+2  𝑠𝑖𝑛𝐴  𝑐𝑜𝑠𝑒𝑐𝐴+𝑐𝑜𝑠2𝐴+𝑠𝑒𝑐2𝐴+2  𝑐𝑜𝑠𝐴  𝑠𝑒𝑐𝐴=𝑠𝑖𝑛2𝐴+𝑐𝑜𝑠2𝐴+𝑐𝑜𝑠𝑒𝑐2𝐴+2  𝑠𝑖𝑛𝐴  1𝑠𝑖𝑛𝐴+𝑠𝑒𝑐2𝐴+2  𝑐𝑜𝑠𝐴  1𝑐𝑜𝑠𝐴=1+𝑐𝑜𝑠𝑒𝑐2𝐴+2+𝑠𝑒𝑐2𝐴+2=5+𝑐𝑜𝑠𝑒𝑐2𝐴+𝑠𝑒𝑐2𝐴=5+1+𝑐𝑜𝑡2𝐴+1+𝑡𝑎𝑛2𝐴=7+𝑡𝑎𝑛2𝐴+𝑐𝑜𝑡2𝐴=𝑅𝐻𝑆

(𝑖𝑥)  (𝑐𝑜𝑠𝑒𝑐𝐴–𝑠𝑖𝑛𝐴)(𝑠𝑒𝑐𝐴–𝑐𝑜𝑠𝐴)=1𝑡𝑎𝑛𝐴+𝑐𝑜𝑡𝐴

[Hint : Simplify LHS and RHS separately]

Convert LHS and RHS in terms of sin and cos.

𝐿𝐻𝑆=(𝑐𝑜𝑠𝑒𝑐𝐴–𝑠𝑖𝑛𝐴)(𝑠𝑒𝑐𝐴–𝑐𝑜𝑠𝐴)=(1𝑠𝑖𝑛𝐴–𝑠𝑖𝑛𝐴)(1𝑐𝑜𝑠𝐴–𝑐𝑜𝑠𝐴)=(1−𝑠𝑖𝑛2𝐴𝑠𝑖𝑛𝐴)(1−𝑐𝑜𝑠2𝐴𝑐𝑜𝑠𝐴)=(𝑐𝑜𝑠2𝐴𝑠𝑖𝑛𝐴)(𝑠𝑖𝑛2𝐴𝑐𝑜𝑠𝐴)=𝑠𝑖𝑛𝐴  𝑐𝑜𝑠𝐴  …  …(1)

𝑅𝐻𝑆=1𝑡𝑎𝑛𝐴+𝑐𝑜𝑡𝐴=1𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴+𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐴=1𝑠𝑖𝑛2𝐴+𝑐𝑜𝑠2𝐴𝑠𝑖𝑛𝐴  𝑐𝑜𝑠𝐴=11𝑠𝑖𝑛𝐴  𝑐𝑜𝑠𝐴=𝑠𝑖𝑛𝐴  𝑐𝑜𝑠𝐴  …  …(2)

Frim (1) and (2)

LHS = RHS

(𝑥)  1+𝑡𝑎𝑛2𝐴1+𝑐𝑜𝑡2𝐴=(1−𝑡𝑎𝑛𝐴1−𝑐𝑜𝑡𝐴)2=𝑡𝑎𝑛2𝐴

There are three terms that we need to prove equal to each other.  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+𝑡𝑎𝑛2𝐴1+𝑐𝑜𝑡2𝐴=1+𝑠𝑖𝑛2𝐴𝑐𝑜𝑠2𝐴1+𝑐𝑜𝑠2𝐴𝑠𝑖𝑛2𝐴=𝑐𝑜𝑠2𝐴+𝑠𝑖𝑛2𝐴𝑐𝑜𝑠2𝐴𝑠𝑖𝑛2𝐴+𝑐𝑜𝑠2𝐴𝑠𝑖𝑛2𝐴=1𝑐𝑜𝑠2𝐴1𝑠𝑖𝑛2𝐴=1𝑐𝑜𝑠2𝐴×𝑠𝑖𝑛2𝐴1=𝑡𝑎𝑛2𝐴=𝑅𝐻𝑆

𝑀𝑇=(1−𝑡𝑎𝑛𝐴1−𝑐𝑜𝑡𝐴)2=(1−𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴1−𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐴)2=(𝑐𝑜𝑠𝐴–𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐴−𝑐𝑜𝑠𝐴𝑠𝑖𝑛𝐴)2=[𝑐𝑜𝑠𝐴–𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴×𝑠𝑖𝑛𝐴𝑠𝑖𝑛𝐴−𝑐𝑜𝑠𝐴]2=[−(𝑠𝑖𝑛𝐴−𝑐𝑜𝑠𝐴)𝑐𝑜𝑠𝐴×𝑠𝑖𝑛𝐴𝑠𝑖𝑛𝐴−𝑐𝑜𝑠𝐴]2=[−𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴]2=(−𝑡𝑎𝑛𝐴)2=𝑡𝑎𝑛2𝐴=𝑅𝐻𝑆

Recommended Tutorial

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

$\begin{aligned}  \end{aligned}$