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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