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.