Question

$\frac{{\sin }^4\theta -{\cos }^4\theta }{1-{\sin }^2\theta }=$ how much?

• A. $1-{\cot }^2\theta$
• B. $1-{\tan }^2\theta$
• C. ${\tan }^2\theta -1$
• D. ${\cot }^2\theta -1$

Answer 1

Answer: (C)

${\tan }^2\theta -1$

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

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

$\therefore$ $\frac{{\sin }^4\theta -{\cos }^4\theta }{1-{\sin }^2\theta }={\tan }^2\theta -1$.