Question

sin4theta + cos4theta/1-2sin2thetacos2theta=1​

Answer 1

CORRECT QUESTION :-

$\\ \\ \mapsto \sf \: Prove \: That \: \: \dfrac{ \sin ^{4} \theta + \cos ^{4} \theta }{1 - 2 \sin ^{2} \theta \: \cos ^{2} \theta } = 1$

GIVEN :-

$\\ \\ \mapsto \sf \: \dfrac{ \sin ^{4} \theta + \cos ^{4} \theta }{1 - 2 \sin ^{2} \theta \: \cos ^{2} \theta } = 1$

TO FIND :-

• L.H.S = R.H.S

SOLUTION :-

$\\ \\ \mapsto \sf \dfrac{ \sin ^{4} \theta + \cos ^{4} \theta }{1 - 2 \sin ^{2} \theta \: \cos ^{2} \theta } = 1$

➔ sin⁴ϴ + cos⁴ϴ Can be written as (sin²ϴ)² + (cos²ϴ)².

$\\ \\ \mapsto \sf \dfrac{( \sin ^{2} \theta) ^{2} + (\cos ^{2} \theta ) ^{2} }{1 - 2 \sin ^{2} \theta \: \cos ^{2} \theta } = 1$

➔ By using identity :- a² + b² = (a + b)² -2ab.

$\\ \\ \mapsto \sf \dfrac{(\sin ^{2} \theta + \cos ^{2} \theta ) ^{2} \: -2 \sin ^{2} \theta \: \cos ^{2} \theta }{1 - 2 \sin ^{2} \theta \: \cos ^{2} \theta } = 1$

➔ By using identity :- sin²ϴ + cos²ϴ = 1.

$\\ \\ \mapsto \sf \dfrac{ \cancel{1 \: -2 \sin ^{2} \theta \: \cos ^{2} \theta} }{ \cancel{1 - 2 \sin ^{2} \theta \: \cos ^{2} \theta } } = 1 \\ \\ \\ \mapsto \sf \: 1 = 1$

L.H.S = R.H.S

❏ HENCE VERIFIED

ADDITIONAL INFORMATION :-

☄ Some Identities:-

$\mapsto \sf \tan (A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \: \tan B} \\ \\ \\ \mapsto \sf \tan (A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \: \tan B} \\ \\ \\ \mapsto \sf \sin^{2} \theta + \cos^{2} \theta = 1 \\ \\ \\ \mapsto \sf \sin^{2} \theta = 2 \sin \theta \ \cos \theta$