Question

Prove that ${\tan }^2\theta {\cos }^2\theta =1-{\cos }^2\theta$.

Answer 1

LHS = ${\tan }^2\theta {\cos }^2\theta$

$=\frac{{\sin }^2\theta }{{\cos }^2\theta }\times {\cos }^2\theta$

$={\sin }^2\theta$

$=1-{\cos }^2\theta$

$=RHS$