Trigonometric Identities
Class 10th Mathematics Rajasthan Board Solution
- Express all the trigonometric ratios ∠θ in terms of sec θ.
- Express the trigonometric ratios sin θ, sec θ, tan θ in terms of cot θ.…
- cos^2 θ + cos^2 θ.cot^2 θ = cot^2 θ Prove the following with the help of…
- sec θ (1 - sin θ) (sec θ + tan θ) = 1 Prove the following with the help of…
- cosec^2 θ + sec^2 θ = cosec^2 θ sec^2 θ Prove the following with the help of…
- root 1-sintegrate heta /1+sintegrate heta = sectheta -tantheta Prove the…
- root sec^2theta +cosec^2theta = tantheta +cottheta Prove the following with the…
- tanalpha +tanbeta /cotalpha +cotbeta = tanalp anbeta Prove the following with…
- 1+sintegrate heta /costheta = costheta /1+sintegrate heta = 2sectheta Prove the…
- sin^4theta -cos^4theta /sin^2theta -cos^2theta = 1 Prove the following with the…
- cottheta -tantheta = 1-2sin^2theta /sintegrate heta costheta Prove the…
- cos^4 θ + sin^4 θ = 1 - 2 cos^2 θ sin^2 θ Prove the following with the help of…
- (secθ - cosθ) (cotθ + tanθ) = tanθsecθ Prove the following with the help of…
- 1-tan^2alpha /cot^2alpha -1 = tan^2alpha Prove the following with the help of…
- sintegrate heta /1-costheta = 1+costheta /sintegrate heta Prove the following…
- sin^6 θ + cos^6 θ = 1 - 3sin^3 θ cos^2 θ Prove the following with the help of…
- tantheta /1-cottheta + cottheta /1-tantheta = 1+tantheta +cottheta Prove the…
- sinθ (1 + tanθ) + cosθ(1 + cotθ) = cosecθ + secθ Prove the following with the…
- sin^2 θ cosθ + tanθsinθ + cos^3 θ = secθ Prove the following with the help of…
- tantheta /1-cottheta + cottheta /1-tantheta = 1+sectheta +cosectheta Prove the…
- (sin A + cosec A)^2 + (cos A + sec A)^2 = 7 + tan^2 A + cot^2 A Prove the…
- sin^8 θ - cos^8 θ = (sin^2 θ - cos^2 θ) (1- 2sin^2 θ cos^2 θ) Prove the…
- root sectheta +1/sectheta -1 = cottheta +cosectheta Prove the following with…
- (1+cottheta +tantheta) (sintegrate heta -costheta)/sec^3theta -cosec^3theta =…
- sintegrate heta +costheta /sintegrate heta -costheta + sintegrate heta…
- cosa/1-tana + sina/1-cota = sina+cosa Prove the following with the help of…
- (coseca-sina) (seca-cosa) = 1/tana+cota Prove the following with the help of…
- cos^2theta /1-tantheta + sin^3theta /sintegrate heta -costheta = 1+sintegrate…
- If sec θ + tan θ = P, then prove that p^2 - 1/p^2 + 1 = sintegrate heta Prove…
- cosa/cosb = m cosa/sinb = n then prove that (m^2 + n^2) cos^2 B = n^2 Prove the…
- (i) cos37^circle /sin53^circle (ii) cosec32^circle /sec58^circle (iii)…
- Find the values of the following: (i) cosec 25 sec 65 (ii) cot 34 tan 56 (iii)…
- (i) sin 70° sin 20° - cos 20° cos 70° (ii) 2cos67^circle /sin23^circle -…
- (i) (sin35^circle /cos55^circle)^2 + (cos55^circle /sin35^circle)^2 -…
- (i) tan 12° cot 38° cot 52° cot 60° tan 78° (ii) tan 5° tan 25° tan 30° tan 65°…
- Express the following in terms of the trigonometric ratios of angles between 0°…
- sin 65° + cos 25° = 2 cos 25° Prove the following:
- cos70^circle /sin20^circle + cos59^circle /sin31^circle - 8sin^230^circle = 0…
- sin (90^circle - theta) cos (90^circle - theta) = tantheta /1+tan^2theta Prove…
- cos (90^circle - theta) costheta /tantheta +cos^2 (90^circle - theta) = 1 Prove…
- tan (90^circle - theta) cottheta /cosec^2theta -cos^2theta = 0 Prove the…
- cos (90^circle - theta) sin (90^circle - theta)/tan (90^circle - theta) =…
- sintegrate heta cos (90^circle - theta) costheta /sec (90^circle - theta) +…
- If sin 3θ = cos (θ - 6°) where 3θ and (θ - 6°) are acute angles then find the…
- If sec 5θ = cosec (θ - 36°) where 5θ is an acute angle then find the value of…
- If A, B and C are the interior angles of any triangle ABC then prove that tan…
- If cos 2θ = sin 4θ are acute angles then find the value of θ.
Exercise 7.1
Question 1.Express all the trigonometric ratios ∠θ in terms of sec θ.
Answer:
As we know
sin2 θ + cos2 θ = 1
⇒ sin2 θ = 1 – cos2θ
⇒ sin2 θ = 1 – 
⇒ sin θ = 
cos θ = 
using the identity: sec2 θ – tan2 θ = 1
⇒ tan2 θ = sec2 θ – 1
tan θ = 
cosec θ = 
⇒ cosec θ = 
⇒ cosec θ = 
sec θ = sec θ
cot θ = 
Question 2.
Express the trigonometric ratios sin θ, sec θ, tan θ in terms of cot θ.
Answer:
Using the identity: cosec2 θ – cot2 θ = 1
⇒ cosec θ = 
sin θ = 
⇒ sin θ = 
⇒ 
⇒ sec θ = 
⇒ sec θ = 
tan θ = 
Question 3.
Prove the following with the help of identities:
cos2θ + cos2θ.cot2θ = cot2θ
Answer:
Taking L.H.S we get,
cos2 θ + cos2 θ.cot2 θ
⇒ cos2 θ (1 + cot2 θ)
⇒ cos2 θ (1 + 
)
⇒ cos2 θ (
)
Using the identity: 1 + tan2 θ = sec2 θ
⇒ cos2 θ (
) = 
⇒ = cot2 θ
= R.H.S
Hence, proved.
Question 4.
Prove the following with the help of identities:
sec θ (1 – sin θ) (sec θ + tan θ) = 1
Answer:
Taking L.H.S we get,
sec θ (1 – sin θ) (sec θ + tan θ)
⇒ (sec θ – sec θ.sin θ) (sec θ + tan θ)
⇒ (sec θ – tan θ) (sec θ + tan θ)
Using the identity: (a + b) (a – b) = a2 – b2
⇒ sec2 θ – tan2 θ
= 1
= R.H.S
Hence, proved
Note: sin2 θ + cos2 θ = 1
sec2 θ – tan2 θ = 1
cosec2 θ – cot2 θ = 1
Question 5.
Prove the following with the help of identities:
cosec2θ + sec2θ = cosec2θ sec2θ
Answer:
Taking L.H.S we get,
cosec2 θ + sec2 θ
⇒ 
Taking L.C.M we get,
⇒ 
⇒ 
⇒ cosec2 θ.sec2 θ
= R.H.S
Hence, proved
Question 6.
Prove the following with the help of identities:
Answer:
To rationalizing the denominator multiply by the conjugate to both numerator and denominator: (1 – sin θ)
⇒ 
= 
⇒ 
Using, cos2 θ = 1 – sin2 θ
⇒ 
⇒ sec θ – tan θ
= R.H.S
Hence, proved.
Question 7.
Prove the following with the help of identities:
Answer:
As we know,
Sec2 θ = 1 + tan2 θ and cosec2 θ = 1 + cot2 θ
Taking L.H.S we get,
⇒ 
⇒ 
⇒ 
Using the identity: (a + b)2 = a2 + b2 + 2.a.b
⇒ 
⇒ tan θ + cot θ
= R.H.S
Hence, proved
Question 8.
Prove the following with the help of identities:
Answer:
As we know,
cot α = 
and cot β = 
Taking L.H.S we get,
⇒ 
⇒ 
⇒ tan α.tan β
= R.H.S
Hence, proved
Question 9.
Prove the following with the help of identities:
Answer:
L.H.S = 
⇒ 
(Taking L.C.M and common denominator)
⇒ 
⇒ 
(Using the identity sin2 θ + cos2 θ = 1)
⇒ 
(Taking 2 common from numerator)
⇒ 
= 2 sec θ
= R.H.S
Hence, proved.
Question 10.
Prove the following with the help of identities:
Answer:
Taking L.H.S we get,
Using the identity: (a2 – b2) = (a – b)(a + b)
⇒ 
⇒ sin2 θ + cos2 θ (since, sin2θ + cos2θ = 1)
= 1 = R.H.S
Hence, proved.
Question 11.
Prove the following with the help of identities:
Answer:
As we know,
cot θ = 
and tan θ = 
Taking L.H.S we get,
cot θ – tan θ = 
⇒ 
Using: cos2 θ = 1 – sin2 θ
⇒ 
⇒ 
= R.H.S
Hence, proved.
Question 12.
Prove the following with the help of identities:
cos4θ + sin4θ = 1 – 2 cos2θ sin2θ
Answer:
Taking L.H.S we get,
Cos4 θ + sin4 θ
Using (a2 + b2) = (a + b)2 – 2.a.b
Where a = cos2 θ and b = sin2 θ
⇒ (cos2 θ + sin2 θ)2 – 2.cos2 θ.sin2 θ
⇒ 1 – 2.cos2 θ.Sin2 θ (Using: sin2 θ + cos2 θ = 1 )
= R.H.S
Hence, proved.
Question 13.
Prove the following with the help of identities:
(secθ – cosθ) (cotθ + tanθ) = tanθsecθ
Answer:
As we know,
sec θ = 
, cot θ = and tan θ =
Taking L.H.S we get,
(sec θ – cos θ)(cot θ + tan θ)
⇒ (
)(
)
⇒ (
)(
)
⇒ 
(
) (Using: cos2 θ + sin2 θ = 1 and 1 – cos2 θ = sin2 θ)
⇒ 
⇒ tan θ.sec θ
= R.H.S
Hence, proved.
Question 14.
Prove the following with the help of identities:
Answer:
Taking L.H.S we get,
Using: cot α =
⇒ 
⇒ 
( Taking the L.C.M)
= tan2 α = R.H.S
Hence, proved
Question 15.
Prove the following with the help of identities:
Answer:
Taking the L.H.S we get,
Multiplying by (1 + cos θ) to both numerator and denominator,
⇒ 
⇒ 
(Using: (a + b)(a – b) = a2 – b2)
⇒ 
(Using the identity: sin2 θ + cos2 θ = 1)
⇒ 
= R.H.S
Hence, proved.
Question 16.
Prove the following with the help of identities:
sin6θ + cos6θ = 1 – 3sin3θ cos2θ
Answer:
Using the identity: a3 + b3 = (a + b)(a2 + b2 – ab)
Here, a = sin2 θ and b = cos2 θ
⇒ (sin2 θ + cos2 θ)(sin4 θ + cos4 θ – sin2 θ.cos2 θ)
Now, Using: sin2θ + cos2θ = 1 and a2 + b2 = (a + b)2 – 2.a.b
⇒ (1)((sin2 θ + cos2 θ)2 – 2.sin2 θ.cos2 θ – sin2θ.cos2θ )
⇒ ((1)2 – 2.sin2 θ.cos2 θ – sin2 θ.cos2 θ)
⇒ (1 – 3.sin2 θ.cos2 θ)
= R.H.S
Hence, proved.
Question 17.
Prove the following with the help of identities:
Answer:
Taking L.H.S we get,
⇒ 
⇒ 
⇒ 
(Taking ‘–1’ common from denominator)
⇒ 
⇒ 
(Using: cot θ = 
)
⇒ 
(Using the identity: a3 –b3 = (a – b)(a2 + b2 + ab))
⇒ 
⇒ tan θ + cot θ +1
= R.H.S
Hence, proved.
Question 18.
Prove the following with the help of identities:
sinθ (1 + tanθ) + cosθ(1 + cotθ) = cosecθ + secθ
Answer:
Taking L.H.S we get,
sin θ (1 + ) + cos θ(1 + )
⇒ 
Taking L.C.M and using the identity sin2 θ + cos2 θ = 1.
⇒ 
⇒ 
⇒ 
⇒ sec θ + cosec θ
= R.H.S
Hence, proved.
Question 19.
Prove the following with the help of identities:
sin2θ cosθ + tanθsinθ + cos3θ = secθ
Answer:
Taking L.H.S we get,
Rearranging the terms ;
Sin2 θ.cos θ + cos3 θ + tan θ.sin θ
⇒ cos θ(sin2 θ + cos2 θ) + tan θ.sin θ
⇒ cos θ (1) +
⇒ cos θ +
⇒ 
⇒ = sec θ
= R.H.S
Hence, proved.
Question 20.
Prove the following with the help of identities: = 1 + sec θ cosec θ
Answer:
Taking L.H.S we get,
⇒
⇒ 
Taking L.C.M and using: a3 – b3 = (a – b)(a2 + b2 + ab)
⇒ 
⇒ 
⇒ 
⇒ cot θ + tan θ + 1
⇒ +1
⇒ 
+ 1
⇒ cosec θ.sec θ + 1
= R.H.S
Hence, proved.
Question 21.
Prove the following with the help of identities:
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2A + cot2A
Answer:
Taking L.H.S we get,
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
Expanding the terms using: (a + b)2 = a2 + b2 + 2.a.b
⇒ sin2 A + cosec2 A + 2.sin A.cosec A + cos2 A + sec2 A + 2.sec A.cos A
⇒ (sin2 A + cos2 A) + 2.sin A.
+ cosec2 A + sec2 A + 2.cos A.
⇒ 1 + 2 + cosec2 A + sec2 A + 2
⇒ 5 + (1 + cot2 A) + (1 + tan2 A)
⇒ 5 + 1 + 1 + tan2 A + cot2 A
⇒ 7 + tan2 A + cot2 A
= R.H.S
Hence, proved.
Question 22.
Prove the following with the help of identities:
sin8θ – cos8θ = (sin2θ – cos2θ) (1– 2sin2θ cos2θ)
Answer:
Taking L.H.S we get,
Sin8 θ– cos8 θ
Using the formula: a2 – b2 = (a – b)(a + b)
⇒ (sin4 θ – cos4 θ)(sin4 θ + cos4 θ)
⇒ (sin2 θ – cos2 θ)(sin2 θ + cos2 θ)((sin2 θ + cos2 θ)2 – 2.sin2 θ.cos2 θ)
⇒ (sin2 θ – cos2 θ)(1)((1)2 – 2.sin2 θ.cos2 θ) (Using: sin2 θ + cos2 θ = 1.)
⇒ (sin2 θ – cos2 θ)(1 – 2.sin2 θ.cos2 θ)
= R.H.S
Hence, proved
Question 23.
Prove the following with the help of identities:
Answer:
Taking the L.H.S we get,
Using: sec θ = and taking L.C.M
⇒ 
To rationalize the denominator multiply by (1 + cos θ) to both numerator and denominator.
⇒ 
⇒ 
⇒
⇒ cosec θ + cot θ
= R.H.S
Hence, proved.
Question 24.
Prove the following with the help of identities:
Answer:
Taking L.H.S we get,
⇒ 
⇒ 
⇒ 
⇒ 
(using: a3 – b3 = (a – b)(a2 + b2 + ab) and cancelling the term.)
⇒ sin2θ cos2θ
= RHS
Hence, Proved!
Question 25.
Prove the following with the help of identities:
= 
Answer:
Taking L.H.S we get,
Taking L.C.M wee get,
⇒ 
As we know,
(a + b)2 + (a – b)2 = a2 + b2 + 2.a.b + a2 + b2 – 2.a.b
⇒ (a + b)2 + (a – b)2 = 2(a2 + b2)
⇒
⇒ 
= 
And,
⇒ 
⇒ 
Hence, proved.
Question 26.
Prove the following with the help of identities:
Answer:
Taking the L.H.S we get,
⇒ 
⇒ 
⇒ 
⇒ 
(taking the L.C.M)
⇒ 
(using the identity: a2 – b2 = (a + b)(a – b))
⇒ 
= (cos A + sin A)
= R.H.S
Hence, proved.
Question 27.
Prove the following with the help of identities:
Answer:
Taking the L.H.S we get,
(cosec A – sin A)(sec A – cos A)
As cosecθ = 1/sinθ and secθ = 1/cosθ
⇒ 
As sin2θ + cos2 θ = 1
⇒ 
⇒ cos A.sin A
It can be written as
⇒ 
= 
As cotθ = cosθ/sinθ and tanθ = sinθ/cosθ
⇒ 
= R.H.S
Hence, proved.
Question 28.
Prove the following with the help of identities:
Answer:
Taking L.H.S we get,
⇒ 
+ 
⇒ 
⇒ 
⇒ 
⇒ 1 + sin θ.cos θ
= R.H.S
Hence, proved.
Question 29.
Prove the following with the help of identities:
If sec θ + tan θ = P, then prove that 
Answer:
Taking L.H.S we get,
⇒ 
Using the identity: (a + b)2 = a2 + b2 + 2.a.b
And sec2 θ – tan2 θ = 1
⇒ 
⇒ 
⇒ 
⇒ 
⇒ sin θ = R.H.S
Hence, proved.
Question 30.
Prove the following with the help of identities:
then prove that (m2 + n2) cos2B = n2
Answer:
Taking L.H.S we get,
(
)cos2 B
⇒ 
× cos2 B
⇒ 
= 
= n2
= R.H.S
Hence, proved.
Exercise 7.2
Question 1.Find the values of the following:
(i) 
(ii) 
(iii) 
(iv) 
Answer:
Note:
cos (90˚ – θ) = sin θ
Sin (90˚ – θ) = cos θ
(i) ⇒ cos 37˚ = cos (90˚ – 53˚)
⇒ cos 37˚ = sin 53˚
⇒ 
= 
= 1.
= 1(ii) cosec 32˚ = cosec (90˚ – 58˚)
⇒ cosec 32˚ = sec 58˚
⇒ 
= 
= 1.
Therefore, 
= 1
(iii) tan 10˚ = tan (90˚ – 80˚)
⇒ tan 10˚ = cot 80˚
⇒ 
= 
= 1.
Therefore,
= 1(iv) cos 19˚ = cos (90˚ – 71˚)
⇒ cos 19˚ = sin 71˚
⇒ 
= 
= 1.
Therefore, 
= 1
Question 2.
Find the values of the following:
(i) cosec 25° – sec 65°
(ii) cot 34° – tan 56°
(iii) 
(iv) sin θ cos(90° – θ) + cosθ sin(90° – θ)
Answer:
(i) cosec (90˚– 65˚) – sec 65˚
⇒ sec 65˚ – sec 65˚ (since, cosec (90˚ – θ) = sec θ)
= 0
(ii) cot (90˚ – 56˚) – tan 56˚
⇒ tan 56˚ – tan 56˚ (since, cot (90˚ – θ) = tan θ)
= 0.
(iii) 
⇒ 
⇒ 1 – 1
= 0.
(iv)
sin θ cos(90° – θ) + cosθ sin(90° – θ)
⇒ sin θ.sin θ + cos θ.cos θ (since, cosec (90˚ – θ) = sec θ)
⇒ sin2 θ + cos2 θ
⇒ 1.
Question 3.
Find the values of the following:
(i) sin 70° sin 20° – cos 20° cos 70°
(ii) 
Answer:
(i) Sin (90˚ – 20˚)sin 20˚ – cos 20˚ cos 70˚
⇒ cos 20˚ sin (90˚ – 70˚) – cos 20˚cos 70˚
⇒ cos 20˚cos 70˚ – cos 20˚cos 70˚
⇒ 0.
(ii) 
⇒ 
⇒ 2 – 1 – 
⇒
Question 4.
Find the values of the following:
(i) 
(ii) 
Answer:
(i) (
)2 + (
)2 – 2.cos 60˚
⇒ (
)2 + 
– 1
⇒ 1 + 1 – 1
= 1
(ii) 
⇒ 
⇒ 12 + 12
= 2.
Question 5.
Find the values of the following:
(i) tan 12° cot 38° cot 52° cot 60° tan 78°
(ii) tan 5° tan 25° tan 30° tan 65° tan 85°
Answer:
(i) tan 12˚.cot (90˚ – 52˚)cot 52˚cot 60˚tan (90˚–12˚)
⇒ tan 12˚cot 12˚ tan 52˚ cot 52˚ cot 60˚
⇒ tan 12˚×
×tan 52˚×
× 
⇒
(ii) tan 5˚× tan 25˚× ×tan (90˚ – 25˚)tan (90˚–5˚)
⇒ tan 5˚×
××tan 25˚×
⇒ 1×
1
=
Question 6.
Express the following in terms of the trigonometric ratios of angles between 0° and 45°.
(i) sin 81° + sin 71°
(ii) tan 68° + sec 68°
Answer:
(i) sin (90˚ – 9˚) + sin (90˚ – 19˚)
⇒ cos 9˚ + cos 19˚
(ii) tan (90˚ – 22˚) + sec (90˚ – 22˚)
⇒ cot 22˚ + cosec 22˚
Question 7.
Prove the following:
sin 65° + cos 25° = 2 cos 25°
Answer:
sin (90˚ – 25˚) + cos 25˚
⇒ cos 25˚ + cos 25˚
⇒ 2 cos 25˚
= R.H.S (Hence, proved.)
Question 8.
Prove the following:
Answer:
Taking L.H.S we get,
⇒ 
⇒ 
⇒ 1 + 1 –8 sin2 30˚
⇒ 2 – 8 (
)2
⇒ 2 – 8.
⇒ 2 – 2 = 0.
= R.H.S
Hence, proved.
Question 9.
Prove the following:
Answer:
Taking L.H.S we get,
sin (90˚ – θ)cos(90˚ – θ)
⇒ cos θ.sin θ
⇒ 
(Writing 1 as sin2 θ + cos2 θ.)
Dividing by cos2 θ to both numerator and denominator
⇒ 
⇒ 
⇒ 
= R.H.S
Hence, proved.
Question 10.
Prove the following:
Answer:
Taking L.H.S we get,
⇒ 
⇒ cos2 θ + sin2 θ
⇒ 1
= R.H.S
Hence, proved.
Question 11.
Prove the following:
Answer:
Taking L.H.S we get,
⇒ 
⇒ 
– cos2 θ
⇒ cos2 θ – cos2 θ
= 0 = R.H.S
Hence, proved.
Question 12.
Prove the following:
Answer:
Taking L.H.S we get,
⇒ 
⇒ 
⇒ sin2 θ = R.H.S
Hence, proved.
Question 13.
Prove the following:
Answer:
Taking L.H.S we get,
⇒ sin3 θ.cos θ + cos3 θ.sin θ
⇒ sin θ.cos θ (sin2 θ + cos2 θ)
⇒ sin θ.cos θ = R.H.S
Hence, proved.
Question 14.
If sin 3θ = cos (θ – 6°) where 3θ and (θ – 6°) are acute angles then find the value of θ.
Answer:
As we know,
cos θ = sin (90˚ – θ)
⇒ sin 3θ = cos (θ – 6˚)
⇒ sin 3θ = sin (90˚ – (θ – 6˚))
⇒ sin 3θ = sin (90˚ – θ + 6˚)
⇒ sin 3θ = sin (96˚ – θ)
Since, 3θ is acute angle
So, 3θ = 96˚ – θ
⇒ 4θ = 96˚
⇒ θ = 
= 24˚
Question 15.
If sec 5θ = cosec (θ – 36°) where 5θ is an acute angle then find the value of θ.
Answer:
As we know,
cosec θ = sec (90˚ – θ)
⇒ sec 5θ = sec (90˚ – (θ – 36˚))
⇒ sec 5θ = sec (90˚ – θ + 36˚)
⇒ sec 5θ = sec (126˚ – θ)
Since, 5θ ia an acute angle.
⇒ 5θ = 126˚ – θ
⇒ 6θ = 126˚
⇒ θ = 
= 21˚
Question 16.
If A, B and C are the interior angles of any triangle ABC then prove that 
Answer:
Given: A, B and C are the interior angles of triangle.
⇒ A + B + C = 180˚
⇒ B + C = 180˚ – A
Dividing by 2 both sides of above equation,
⇒ 
= 90˚ – 
⇒ Taking Tangent both sides,
⇒ tan () = tan (90˚ – )
⇒ tan () = cot () (tan (90˚ – θ) = cot θ)
Hence, proved.
Question 17.
If cos 2θ = sin 4θ are acute angles then find the value of θ.
Answer:
As given 2θ and 4θ are acute angles.
⇒ cos 2θ = sin 4θ
⇒ sin (90˚ – 2θ) = sin 4θ
⇒ 90˚ – 2θ = 4θ
⇒ 6θ = 90˚
⇒ θ = 
= 15˚