Question

Prove the following identity:​

$\frac{{\text{sec}}^2\theta -{\text{sin}}^2\theta }{{\text{tan}}^2\theta }={\text{cosec}}^2\theta -{\text{cos}}^2\theta$

Answer 1

Proving LHS=RHS

First, we consider Left Hand Side(LHS),
$=\frac{{\text{sec}}^2\theta -{\text{sin}}^2\theta }{{\text{tan}}^2\theta }$
We know that, ${\text{sec}}^2\theta =\frac{1}{{\text{cos}}^2\theta },{\text{tan}}^2\theta =\frac{{\text{sin}}^2\theta }{{\text{cos}}^2\theta }$.
$=\frac{\frac{1}{{\text{cos}}^2\theta }-{\text{sin}}^2\theta }{\frac{{\text{sin}}^2\theta }{{\text{cos}}^2\theta }}$
$=\frac{\frac{1-{\text{sin}}^2\theta {\text{cos}}^2\theta }{{\text{cos}}^2\theta }}{\frac{{\text{sin}}^2\theta }{{\text{cos}}^2\theta }}$
$=\frac{1-{\text{sin}}^2\theta {\text{cos}}^2\theta }{{\text{sin}}^2\theta }$
$=\frac{1}{{\text{sin}}^2\theta }-\frac{{\text{sin}}^2\theta {\text{cos}}^2\theta }{{\text{sin}}^2\theta }$
$={\text{cosec}}^2\theta -{\text{cos}}^2\theta$.
Then, Right Hand Side $={\text{cosec}}^2\theta -{\text{cos}}^2\theta$.
Therefore, LHS=RHS.
Hence, proved.