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.

Trigonometric identity class10th

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.

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