# 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 AWe 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

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