# Class 10 NCERT Solutions- Chapter 8 Introduction To Trigonometry – Exercise 8.4

**Question 1. Express the trigonometric ratios sin A, sec A**,** and tan A in terms of cot A**

**Solution:**

(i)sin AHey! Looking for some great resources suitable for young ones? You've come to the right place. Check out our

self-paced courses designed for students of grades I-XII.Start with topics like

Python, HTML, ML, and learn to make some games and appsall with the help of our expertly designed content! So students worry no more, becauseGeeksforGeeks Schoolis now here!

We know that

cosec

^{2}A = 1 + cot^{2}A1/sin

^{2}A = 1 + cot^{2}Asin

^{2}A = 1/(1 + cot^{2}A)sin A = 1/(1+cot

^{2}A)^{1/2}

(ii) sec Asec

^{2}A = 1 + tan^{2}ASec

^{2}A = 1 + 1/cot^{2}Asec

^{2}A = (cot^{2}A + 1) / cot^{2}Asec A = (cot

^{2}A + 1)^{1/2}/ cot A

(iii) tan Atan A = 1 / cot A

tan A = cot

^{-1}A

**Question 2. Write all the other trigonometric ratios of ∠A in terms of sec A**.

**Solution:**

(i) cos Acos A = 1/sec A

(ii) sin AWe know that

sin

^{2}A = 1 – cos^{2}AAlso , cos

^{2}A = 1 / sec^{2}Asin

^{2}A = 1 – 1 / sec^{2}Asin

^{2}A = (sec^{2}A – 1) / sec^{2}Asin A = (sec

^{2}A – 1)^{1/2}/ sec A

(iii) tan AWe know that

tan

^{2}A + 1 = sec^{2}Atan A = (sec

^{2}A – 1)½

(iv) cosec AWe know

cosec A = 1/ sinA

cosec A = sec A / (sec

^{2}A – 1)½

(v) cot AWe know

cot A = cos A / sin A

cot A = (1/sec A) / ((sec

^{2}A – 1)^{1/2}/ sec A)cot A = 1 / (sec

^{2}A – 1)^{1/2}

**Question 3. Evaluate: **

**(i) (sin ^{2} 63° + sin^{2} 27°)/(cos^{2} 17° + cos^{2} 73°)**

**(ii) sin 25° cos 65° + cos 25° sin 65°**

(i)([sin(90-27)]^{2}+ sin^{2}27) / ([cos(90-73)]^{2}+ cos^{2}73)We know that

sin(90-x) = cos x

cos(90-x) = sin x(cos

^{2}(27) + sin^{2}27) / (sin^{2}(73) + cos^{2}73)Using

sin

^{2}A + cos^{2}A = 11/1 = 1

(ii)[sin 25 * cos(90-25)] + [cos 25 * sin(90-25)]Using

sin(90-x) = cos x

cos(90-x) = sin x= [sin 25 * sin 25] + [cos 25 * cos 25]

= sin

^{2}25 + cos^{2}25= 1

**Question 4. Choose the correct option. Justify your choice.**

**Solution:**

(i) 9 sec^{2 }A – 9 tan^{2}A(

A) 1 (B) 9 (C) 8 (D) 0Using sec

^{2}A – tan^{2}A = 19 (sec

^{2}A – tan^{2}A ) = 9(1)

Ans (B)

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

(A) 0 (B) 1 (C) 2 (D) –1Simplifying all ratios

= (1 + sinθ/cosθ + 1/cosθ) (1 + cosθ/sinθ – 1/sinθ)

= ((cosθ + sinθ + 1)/ cosθ) ((sinθ + cosθ – 1 )/sinθ)

= ((cosθ + sinθ)

^{2}– 1) / (sinθ cosθ)= (1 + 2*cosθ*sinθ – 1) / (sinθ cosθ)

= 2

Ans (C)

(iii) (sec A + tan A) * (1 – sin A)

(A) sec A (B) sin A (C) cosec A (D) cos ASimplifying sec A and tan A

= (1/cos A + sin A/cos A)*(1 – sin A)

= ((1 + sin A)/cos A)*(1 – sin A)

= (1 – sin

^{2}A)/cos A= cos

^{2}A / cos A= cos A

Ans (D)

(iv) (1 + tan^{2}A) / (1 + cot^{2}A)

(A) sec^{2}A (B) –1 (C) cot^{2}A (D) tan^{2}ASimplifying tan A and cot A

= (1 + (sin

^{2}A / cos^{2}A)) / (1 + (cos^{2}A / sin^{2}A))= ((cos

^{2}A + sin^{2}A) / cos^{2}A) / ((cos^{2}A + sin^{2}A) / sin^{2}A)= sin

^{2}A / cos^{2}A= tan

^{2}A

Ans (D)

**Question 5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.**

**Solution:**

(i) (cosec θ – cot θ)^{2 }= (1 – cosθ) / (1 + cosθ)Solving LHS

Simplifying cosec θ and cot θ

= (1-cos θ)

^{2}/ sin^{2}θ= (1-cos θ)

^{2}/ (1-cos^{2}θ)Using a

^{2}– b^{2}= (a+b)*(a-b)= (1-cos θ)

^{2}/ [(1-cos θ)*(1+cos θ)]= (1-cos θ) / (1+cos θ) = RHS

Hence Proved

(ii) (cos A / (1+sin A) + ((1+sin A) / cos A) = 2 sec ASolving LHS

Taking LCM

= (cos

^{2}A + (1+sin A)^{2}) / ((1+sin A) cos A)= (cos

^{2}A + 1 + sin^{2}A + 2 sin A ) / ((1 + sin A)*cos A)Using sin

^{2}A + cos^{2}A = 1= (2 + 2*sin A) / ((1+sin A)*cos A)

= (2*(1 + sin A)) / ((1 + sin A)*cos A)

= 2 / cos A

= 2 sec A = RHS

Hence Proved

(iii) (tan θ / (1 – cot θ)) + (cot θ / (1 – tan θ)) = 1 + sec θ*cosec θSolving LHS

Changing tan θ and cot θ in terms of sin θ and cos θ and simplifying

= ((sin

^{2}θ) / (cos θ *(sin θ-cos θ))) + ((cos^{2}θ ) / (sin θ *(sin θ-cos θ)))= (1 / (sin θ-cos θ)) * [(sin

^{3}θ – cos^{3}θ) / (sin θ * cos θ)]= (1 / (sin θ – cos θ)) * [ ((sin θ – cos θ) * ( sin

^{2}θ + cos^{2}θ + sin θ * cos θ ))/(sin θ *cos θ)]= (1+sin θ*cos θ) / (sin θ*cos θ)

= sec θ*cosec θ + 1 = RHS

Hence Proved

(iv) (1 + sec A) / sec A = sin^{2}A / (1 – cos A)Solving LHS

= cos A + 1

Solving RHS

= (1 – cos

^{2}A) / (1 – cos A)= (1 – cos A) * (1 + cos A) / (1 – cos A)

= 1 + cos A = RHS

Hence Proved

(v) (cos A – sin A + 1) / (cos A + sin A – 1) = cosec A + cot A using the identity cosec^{2}A = 1 + cot^{2}ASolving LHS

Multiplying numerator and denominator by (cot A – 1 + cosec A)

= (cot

^{2}A + 1 + cosec^{2}A – 2*cot A – 2*cosec A + 2*cot A*cosec A) / (cot^{2}A – (1 + cosec^{2}A – 2*cosec A))= (2*cosec

^{2}A – 2*cot A – 2*cosec A + 2*cot A*cosec A) / (cot^{2}A – 1 – cosec^{2}A + 2*cosec A)= (2* cosec A *(cosec A + cot A) – 2*(cosec A + cot A)) / (cot

^{2}A – 1 – cosec^{2}A + 2*cosec A)= ((cosec A + cot A) * (2*cosec A – 2 )) / (2*cosec A – 2)

= cosec A + cot A = RHS

Hence Proved

(vi) [(1 + sin A) / (1 – sin A)]^{½}= sec A + tan ASolving LHS

Multiplying numerator and denominator by (1+sinA)

= [((1 + sin A)*(1 + sin A)) / ((1 – sin A)*(1 + sin A))]

^{½}= (1 + sin A) / (1 – sin

^{2}A)^{½}= (1 + sin A) / (cos

^{2}A)^{1/2}= (1 + sin A) / (cos A)

= sec A + tan A = RHS

Hence Proved

(vii) (sin θ – 2 sin^{3}θ) / (2 cos^{3}θ – cos θ) = tan θSolving LHS

= (sin θ * (1 – 2*sin

^{2}θ)) / (cos θ * (2*cos^{2}θ – 1))= (sin θ * (1 – 2*sin

^{2}θ )) / (cos θ * (2*(1 – sin^{2}θ) – 1))= (sin θ *(1 – 2*sin

^{2}θ)) / (cos θ * (1 – 2*sin^{2}θ))= tan θ = RHS

Hence Proved

(viii) (sin A + cosec A)^{2 }+ (cos A + sec A)^{2}= 7 + tan^{2}A + cot^{2}ASolving LHS

= sin

^{2}A + cosec^{2}A + 2*sin A *cosec A + cos^{2}A + sec^{2}A + 2*cos A *sec AWe know that cosec A = 1 / sin A

= 1 + 1 + cot

^{2}A + 1 + tan^{2}A + 2 + 2= 7 + tan

^{2}A + cot^{2}A = RHS

Hence Proved

(ix) (cosec A – sin A)*(sec A – cos A) = 1 / (tan A + cot A)Solving LHS

= ((1/sin A) – sin A) * ((1/cos A) – cos A)

= ((1 – sin

^{2}A) / sin A) * ((1 – cos^{2}A) / cos A)= (cos

^{2}A * sin^{2}A) / (sin A * cos A)= sin A * cos A

Solving RHS

Simplifying tan A and cot A

= (sin A * cos A) / ( sin

^{2}A + cos^{2}A)= sin A * cos A = RHS

Hence Proved

(x) (1 + tan^{2}A) / (1 + cot^{2}A ) = [(1 – tan A) / (1 – cot A)]^{2}= tan^{2}ASolving LHS

Changing cot A = 1 / tan A

= (tan

^{2}A * (1 + tan^{2}A)) / (1 + tan^{2}A) = tan^{2}A = RHS= [(1 – tan A) / (1 – cot A)]

^{2}= (1 + tan^{2}A – 2*tan A) / (1 + cot^{2}A – 2*cot A)= (sec

^{2}A – 2*tan A) / (cosec^{2}A – 2*cot A)Solving this we get

= tan

^{2}A

Hence Proved