the value of 1- sin2Q/1+cosQ
Answers
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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