Question

Prove that: $[1+\frac{1}{{\tan }^2\theta }][1+\frac{1}{{\cot }^2\theta }]=\frac{1}{{\sin }^2\theta -{\sin }^4\theta }$

Answer 1

$[1+\frac{1}{{\tan }^2\theta }][1+\frac{1}{{\cot }^2\theta }]=\frac{1}{{\sin }^2\theta -{\sin }^4\theta }$

Taking L.H.S., we have-

$\Rightarrow [1+\frac{1}{{\tan }^2\theta }][1+\frac{1}{{\cot }^2\theta }]$

$\Rightarrow =\left[\frac{{\tan }^2\theta +1}{{\tan }^2\theta }\right]\left[\frac{{\cot }^2\theta +1}{{\cot }^2\theta }\right]$

$\Rightarrow =\left[\frac{{\sec }^2\theta }{{\tan }^2\theta }\right]\left[\frac{{\csc }^2\theta }{{\cot }^2\theta }\right](\because 1+{\tan }^2\theta ={\sec }^2\theta \&1+{\cot }^2\theta ={\csc }^2\theta )$

$\Rightarrow =\left[\frac{\left(\frac{1}{{\cos }^2\theta }\right)}{\left(\frac{{\sin }^2\theta }{{\cos }^2\theta }\right)}\right]\left[\frac{\left(\frac{1}{{\sin }^2\theta }\right)}{\left(\frac{{\cos }^2\theta }{{\sin }^2\theta }\right)}\right]$

$\Rightarrow =\left[\frac{1}{{\sin }^2\theta }\right]\left[\frac{1}{{\cos }^2\theta }\right]$

$\Rightarrow =\left[\frac{1}{{\sin }^2\theta }\right]\left[\frac{1}{1-{\sin }^2\theta }\right](\because {\sin }^2\theta +{\cos }^2\theta =1)$

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

$\Rightarrow =\frac{1}{{\sin }^2\theta -{\sin }^4\theta }$

$=R.H.S.$

Hence, it is proved that L.H.S. = R.H.S., i.e., $\{[1+\frac{1}{{\tan }^2\theta }][1+\frac{1}{{\cot }^2\theta }]=\frac{1}{{\sin }^2\theta -{\sin }^4\theta }\}$.