NCERT Trigonometry Class 10 Exercise 8.4 Solutions

Trigonometric identities are very important during solving Questions. Here are the following identities which are used to solve or prove the questions.

I. sin² θ + sin² θ = 1

(a)  sin²θ = 1 – cos² θ

(b) cos²θ = 1 – sin²θ

II.  1 + tan²θ = sec²θ

(a) tan²θ = sec²θ – 1

(b) sec²θ – tan²θ = 1

III. 1 + cot²θ = cosec²θ

(a) cot²θ =  cosec²θ – 1

(b) cosec²θ – cot²θ = 1

1.  sinθ = 1/cosecθ                         ;      cosecθ = 1/sinθ

2. cosθ = 1/secθ                             ;      secθ = 1/cosθ

3. a² – b² = (a + b) (a – b)

4. a³ – b³ = (a + b) (a² – ab + b²)

Q1. Express each of the following ratios sin A, cos A, and tan A in terms of cot A.

solution:

We know that

1 + cot²θ = cosec²θ

1/cosec²A = 1/1+cot²A

=> sin²A = 1/1+cot²A

We know that

1 + tan²A = sec²A

=> 1 + 1/cot²A = sec²A

=> (cot²A + 1)/cot² A = sec² A  {after taking LCM}

=> √(cot² A + 1)/cot² A = sec A     { square root of both side}

cosθ = 1/sec A

=> cos A = 1/√(cot² A + 1)

tan θ = 1/cot A

Q2.Write all the trigonometric ratios of ∠A in terms of sec A.

solution:

(i) cos A = 1/sec A

(ii)  sin² A = 1 – cos² A

=>  sinA = √ 1 – cos² A

=>  sinθ = √ 1 – 1/sec² A

=>sinA = √ [ (sec² A – 1)/sec² A]

=>sinA = √ (sec²A – 1)/sec A

(iii) 1 + tan²A = sec² A

=> tan² A = sec² A – 1

=> tan A = √sec² A – 1

(iv) cot A = 1/tan A

=> cot A = 1/√sec² A – 1

(v) cosec A = 1/sin A

=> cosec A = 1/√ (sec² A – 1)/sec A

Q3. Evaluate :

(i) sin² 63° + sin² 27º / cos² 17º + cos² 73º

Solution:

sin² 63° + sin² 27º / cos² 17º + cos² 73º

= sin² 63°+ sin² (90º – 63º)  / cos²(90º – 73º) + cos² 73º       {  sin( 90° – θ) = cos θ; cos( 90° – θ) = sin θ  }

=  sin² 63° + cos² 63º   / sin² 73º+ cos² 73º

= 1/1    { After applying sin² θ + sin² θ = 1}

= 1

(ii) sin 25º cos65º + cos25º sin65º

Solution:

sin 25º cos65º + cos25º sin65º

= sin25º cos( 90° – 25º ) + cos25º sin ( 90°- 25º)

= sin25º sin25º +    cos25º cos25º

= sin²25°  + cos²25º

= 1

Q4. Choose the correct option. Justify your choice.

(i) 9 sec² A – 9 tan² A =

(A) 1

(B) 9

(C) 8

(D) 0

Ans (B)

Explanation:

9 sec² A – 9 tan² A = 9 ( sec² A – tan² A)

= 9 (1)

= 9

(ii)  ( 1 + tan θ + secθ ) ( 1 + cot θ – cosec θ) =

(A) 0

(B) 1

(C) 2

(D) – 1

Ans (c) 2

Explanation:

( 1 + tan θ + secθ ) ( 1 + cot θ – cosec θ)

= ( 1 + sin θ / cosθ + 1 / cos θ ) ( 1 + cos θ/sin θ – 1/sin θ)

= [ ( cos θ + sin θ + 1) /cos θ ] [ (sin θ + cos θ – 1 )/sin θ]

= [ { (cosθ+sinθ)+1}{(sinθ+cosθ)-1}]/cos θ sin θ

= [ (cos θ + sin θ)² – (1)²]/cos θ sin θ

= [ cos² θ +sin² θ + 2cos θ sin θ – 1 ]/ cos θ sin θ

= [ 1 + 2cos θ sin θ – 1 ]/cos θ sin θ

= [ 2cos θ sin θ ]/cosθsinθ

= 2

(iii) ( secA + tanA ) ( 1 – sin A) =

(A) sec A

(B) sin A

(C) cosec A

(D) cos A

Explanation:

( sec A + tan A) ( 1 – sin A) = (sec A + tan A ) ( 1 – sin A)

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

= ( 1 + sinA )( 1 – sin A)/cos A

= (1)² – sin² A / cos A

=  1 – sin² A  / cos A         { sin² θ + cos² θ = 1}

= cos² A/cos A

= cos A

Ans (D) cos A

(iv)  1 + tan² A / 1 + cot² A =

(A) sec²A                                           (B) -1                                        (C) cot² A                                                  (D) tan² A

Explanation:

1 + tan² A / 1 + cot² A

= sec² A/cosec² A

= ( 1/cos² A ) ÷ (1/sin² A)

= (1/cos² A ) × (sin² A/ 1)

= sin² A /cos² A

= tan² A

Ans  (D)

Q5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) (cosec θ – cot θ)² = 1 – cos θ /1 + cos θ

LHS = (cosec θ – cot θ)²

=> LHS = ( 1/sin θ – cos θ/sin θ)²

=> LHS = (1 – cos θ) ²/sin² θ

=> LHS = ( 1 – cosθ)² /1 -cos² θ

=> LHS = ( 1 – cos θ) ( 1 – cos θ) / ( 1 + cos θ) ( 1 – cos θ)

=> LHS = (1 – cos θ)/(1 + cos θ)

=> LHS = RHS

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

LHS = cos A /1 + sin A  + (1 + sin A)/cos A

=> LHS = cos² A + ( 1 + sin A)² / (1 + sin A) cos A

=> LHS =[cos² A + 1 + 2sin A +sin² A ]/ (1 + sin A) cos A

=> LHS =[1+1+2sinA ]/(1+sinA)cosA

=> LHS =[2+2sinA]/(1+sinA)cosA

=> LHS = 2[1+sinA]/(1+sinA)cosA

=> LHS = 2 × 1/cos A

=> LHS = 2 × sec A                  { secθ=1/cosθ }

=> LHS = 2sec A

=> LHS = RHS

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

LHS = tan θ / 1 – cot θ + cot θ/1 – tan θ

=> LHS = (sin θ/cos θ)/[ 1 – (cos θ /sin θ) ] +( cos θ/sin θ)/[1 – (sin θ/cos θ) ]

Taking LCM in the denominator

=> LHS = (sin θ/cos θ) ÷ [sin θ – cos θ / sin θ ] +  (cos θ/sin θ‎) ÷ [ ( cos θ – sin θ) / cosθ ]

=> LHS = ( sin θ/cos θ) × [ sin θ /sin θ – cos θ ] + (cos θ/sin θ‎) ×[ cos θ /(cos θ – sin θ)]

=> LHS = sin² θ/cos θ (sin θ – cos θ )   + cos² θ/sin θ(cos θ – sin θ)

=> LHS = sin² θ/cos θ(sin θ-cos θ) + cos² θ/- sin θ(- cos θ + sin θ)     {takin – as a common in denominator}

=> LHS= sin² θ/cos θ(sin θ – cos θ) – cos² θ/sin θ (sin θ – cos θ)

Again taking LCM cosθsinθ(sinθ-cosθ)

=> L.H.S. = sin³ θ – cos³ θ / cos θ sin θ(sin θ – cos θ)

After applying identity a³ – b³ = (a – b)(a² – ab + b²) in numerator

=> L.H.S. = (sin θ – cos θ) (sin² θ – sin θ cos θ + cos² θ)/cos θ sin θ (sin θ – cos θ)

after the cancellation of (sin θ -cos θ) we get

=> L.H.S.= ( sin² θ – sin θ cos θ + cos² θ)/cos θ sin θ

After applying identity sin² θ+cos² θ = 1

=> L.H.S. = (1 + sin θ cos θ )/cos θ sin θ

=> LHS = 1/cos θsin θ + sin θ cos θ)/cos θ sin θ

=> L.H.S. = 1/cos θ sin θ + sin θ cos θ )/cos θ sin θ

=> L.H.S. = sec θ cosec θ + 1

=> LHS = RHS                                          Hence proved

 

(iv) 1 + sec A / sec A = sin² A / 1 – cos A

L.H.S. = 1 + sec A/sec A

=> L.H.S. = 1 + (1/cos θ) /( 1/cos θ)

=> L.H.S. = [(cos θ + 1)/cos θ ]/(1/cos θ)

=> L.H.S. = cosθ + 1

R.H.S. = sin² A / 1 – cos A

In the Numerator applying identity sin²θ+cos²θ=1

R.H.S. = 1 – cos² A / 1 – cos A

=> R.H.S. = ( 1 + cos A) (1 – cos A)/ 1 – cos A

=> R.H.S. = (1 + cos A)

L.H.S. = R.H.S

(v) cos A – sin A + 1 / cos A + sin A – 1 = cosec A + cot A  , using the identity cosec² A = 1 + cot² A

L.H.S. =

cos A – sin A + 1 / cos A + sin A – 1

Dividing in numerator and denominator by sinθ

=> L.H.S. =  [(cos A – sin A + 1)/sin θ]/[( cos A + sin A – 1 )/sin θ]

=> L.H.S. = [ (cos A/sin θ)-(sinA/sinθ)+(1/sinθ)]/[( cosA/sinθ)+(sinA/sinθ)-(1/sinθ)]

=>L.H.S.= [cotA-1+cosecA]/[ cotA+1-cosecA]

Rationalize the denominator (cotA+1)-cosecA

=> L.H.S.= [(cotA-1+cosecA)/(cotA+1-cosecA)]× (cotA+1)+cosecA / (cotA+1)+cosecA

=> L.H.S.=[{(cotA+cosecA)-1}{(cotA+cosecA)+1}]/[(cotA+1)²-cosec²A]

=> L.H.S.=[(cotA+cosecA)²-(1)²]/[cot²A+2cotA+1-cosec²A]

=> L.H.S. = [cot²A + 2cotAcosecA + cosec²A – 1]/[cot²A + 2cotA + 1 – cosec²A]

Applying identity cosec²A = 1 + cot²A

=> L.H.S. = [cot² A + 2cot A cosec A + (1 + cot²A) – 1]/[ cot²A + 2cotA + 1 – (1 + cot² A)]

=> L.H.S. = [cot² A + 2cot A cosec A + 1 + cot² A – 1]/[cot² A + 2cotA + 1 – 1 – cot² A]

=> L.H.S. = [ cot² A + 2cot A cosec A + 1 + cot² A – 1]/[ cot²A + 2cot A + 1 – 1 – cot² A]

=> L.H.S. = [2cot²A + 2cot A cosec A]/[2cot A]

=> L.H.S. = 2cot A [cot A + cosec A ]/2cot A

=> L.H.S. = cot A + cosec A

=> L.H.S. = R.H.S.

(vi) √(1 + sin A) / 1 – sin A = sec A + tan A

L.H.S. = √(1 + sin A) / (1 – sin A)

Rationalize the denominator 1 – sin A

=>L.H.S. = √1 + sin A / 1 – sin A

=>L.H.S. = √[(1 + sin A / 1 – sin A)]×[(1 + sin A)/(1 + sin A)]

by using identity a²-b²=(a+b)(a-b)

=>L.H.S. =  √[(1+sin A )²/( 1)²-(sin A)²]

=>L.H.S. = √[(1 + sin A )²/ 1 – sin² A]

By Applying identity sin² θ + cos² θ = 1

=> L.H.S. = √[(1 + sin A )²/ cos² θ]

After taking the square root

=> L.H.S = (1 + sin A)/cosθ

=> L.H.S. =1/cos θ + sin θ/cos θ

=> L.H.S. = sec θ + tan θ     {  sec θ = 1/cos θ; tan θ = sinθ/cosθ  }

=> L.H.S. = R.H.S.

(vii) sin θ-2sin³ θ /2cos³ θ – cos θ = tan θ

L.H.S. = sin θ – 2sin³ θ /2cos³ θ – cos θ

=> L.H.S. = sin θ – 2sin³ θ /2cos³ θ – cos θ

=> L.H.S. = sin θ(1 – 2sin² θ)/cos θ(2cos² θ – 1)

by applying identity sin²θ+cos²θ=1=>cos²θ=1-sin²θ

=> L.H.S. = sinθ(1 – 2sin² θ)/cos θ[2(1-sin² θ) – 1]

=> L.H.S. = sin θ(1 – 2sin² θ)/cos θ[2(1 – sin² θ) – 1]

=> L.H.S. = sin θ(1 – 2sin² θ)/cos θ[2 – 2sin² θ – 1]

=> L.H.S. = sin θ(1 – 2sin² θ)/cos θ[2 – 2sin² θ – 1]

=> L.H.S. = sin θ(1 – 2sin² θ)/cos θ[1 – 2sin² θ]

=> L.H.S. = sin θ/cos θ

=> L.H.S. = tan θ

=> L.H.S. = R.H.S                                                                        Hence Proved

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A

=> L.H.S. = (sin A + cosec A)² + (cos A + sec A)²

=> L.H.S. = sin² A +2sin A cosec A + cosec² A + cos² A + 2cos A sec A + sec² A

=> L.H.S. = sin² A + cos² A + 2sin A cosec A + cosec² A + 2cos A sec A + sec² A

=> L.H.S. = 1 + 2sin A × 1/sin A + cose² A + 2cos A × 1/cos A + sec² A     {sin² θ + cos² θ = 1}

=> L.H.S. = 1 + 2 + 1 + cot² A + 2 + 1+ tan² A     {after applying  1 + cot² θ = cosec² θ ; 1+ tan² θ = sec² θ }

=> L.H.S. = 7 + cot² A + tan² A

=> L.H.S. = R.H.S.          Hence Proved

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

L.H.S. =( cosec A – sin A) (sec A – cos A)

L.H.S. = (1/sin A – sin A) (1/cos A – cos A)

=> L.H.S. = [(1 – sin² A)/sin A][(1 – cos² A)/cos A]

applying identity sin² θ + cos² θ = 1

=> L.H.S. = [(cos² A)/sin A][(sin² A)/cos A]

=> L.H.S. = [cos² A sin² A]/sin A cos A

=> L.H.S. = cosA sin A

=> L.H.S.

R.H.S. = 1 / tan A + cot A

R.H.S. = 1/(sin A/cos A) + (cos A/sin A)

=> R.H.S. = 1  /[(sin² A + cos² B)/cos A sin A]

Applying identity sin² θ + cos² θ = 1

=> R.H.S. = 1/[(1)/ cos A sin A]

=> R.H.S. = cos A sin A

L.H.S. = R.H.S.

(x) (1 + tan² A  / 1+cot² A) = ( 1 – tan A  /  1 – cot A )²  =  tan² A

L.H.S.= ( 1 + tan² A  /  1 + cot² A  )

=> L.H.S. = ( 1 + tan² A  / 1 + cot² A  )

by using identities : 1 + tan² θ = sec² θ  ;  1 + cot² θ = cosec² θ

L.H.S.= ( 1 + tan² A  / 1 + cot² A )

=>L.H.S. = sec² A / cosec² A

=>L.H.S. = (1/cos² A)/(1 / sin² A)

=>L.H.S.=(sin² A/cos² A)

=>L.H.S. = tan² A

=>L.H.S. = R.H.S.

( 1 – tan A  /  1 – cot A )²

= [ {1 – (sin A/cos A)} / { 1 – (cos A/sin A)} ]²

= [{(cos A – sin A)/cos A}/{(sin A – cos A)/sin A}]²

= [{(cos A – sin A)/cos A}  /  {(sin A -cos A)/sin A}]²

= [{(cos A – sin A)/cos A} / {- (cos A – sin A)/sin A}]²

= [(cos A – sin A)×sin A / – (cos A – sin A)× cos A]²

= [sin A/- cos A]²

= [- tan A]²

= tan² A

L.H.S. = R.H.S.