Question

do???????

indian navy 2019​

Answer 1

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