Question

the value of 1- sin2Q/1+cosQ

Answer 1

Step-by-step explanation:

The value of \cos^{2} \theta + \cos^{4} \thetacos

2

θ+cos

4

θ = 1

Step-by-step explanation:

Given that

\ sin \theta + \sin^{2} \theta = 1 sinθ+sin

2

θ=1

\sin \theta = 1 - \sin^{2}\thetasinθ=1−sin

2

θ

we know that 1 - \sin^{2} \theta = \cos ^{2}\theta1−sin

2

θ=cos

2

θ

\sin\theta = \cos^{2} \thetasinθ=cos

2

θ ------- (1)

Thus \cos^{2} \theta + \cos^{4} \thetacos

2

θ+cos

4

θ may be written as

\cos^{2} \theta + (\cos^{2} \theta}) ^{2} ------ (2)

Put the value of cos^{2} \thetacos

2

θ in above equation. the above equation becomes,

\sin\theta + sin^{2} \thetasinθ+sin

2

θ ------ (3)

We know that \sin\theta + sin^{2} \thetasinθ+sin

2

θ = 1 thus

⇒ \cos^{2} \theta + \cos^{4} \thetacos

2

θ+cos

4

θ = 1

Thus the value of \cos^{2} \theta + \cos^{4} \thetacos

2

θ+cos

4

θ = 1

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