Question

Simplify the expression-
$(cose{c}^2\theta -1)\times (se{c}^2\theta -1)$

• A. $co{t}^2\theta$
• B. $ta{n}^2\theta$
• C. $1$
• D. $0$

Answer 1

The correct option is C $1$

We know that $cose{c}^2\theta -1=co{t}^2\theta$ and $se{c}^2\theta -1=ta{n}^2\theta$
$\therefore (cose{c}^2\theta -1)\times (se{c}^2\theta -1)$
$=co{t}^2\theta \times ta{n}^2\theta$
$=co{t}^2\theta \times \frac{1}{co{t}^2\theta }$
$=1$.