Math, asked by jayprakashbhaskar0, 10 months ago

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indian navy 2019​

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Answered by AdorableMe
27

Step-by-step explanation:

\frac{1}{sin^6\theta + cos^6\theta} \\\\=\frac{1}{(sin^2\theta)^3+(cos^2\theta)^3} \\\\=\frac{1}{(sin^2\theta+cos^2\theta)(sin^4\theta+cos^4\theta-sin^2\theta.cos^2\theta)}

=\frac{1}{(sin^2\theta-cos^2\theta)^2}\\\\\\ =\frac{1}{(sin^2\theta-cos^2\theta)(sin^2\theta-cos^2\theta)}  

All identities used:

[a³+b³= (a+b)(a²+b²-ab)]

(sin²θ-cos²θ)² = (sin⁴θ+cos⁴θ-sin²θ.cos²θ)

sin²θ+cos²θ=1

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