Question

$se{c}^2\theta +cose{c}^2\theta$ is equal to

• A. $se{c}^2\theta .{\cot }^2\theta$
• B. $se{c}^2\theta .ta{n}^2\theta$
• C. $cose{c}^2\theta .co{t}^2\theta$
• D. $se{c}^2\theta .cose{c}^2\theta$

Answer 1

Answer: (D)

$se{c}^2\theta .cose{c}^2\theta$

We know that $se{c}^2\theta =\frac{1}{co{s}^2\theta }$ and $cose{c}^2\theta =\frac{1}{si{n}^2\theta }$

$\therefore \frac{1}{co{s}^2\theta }+\frac{1}{si{n}^2\theta }=\frac{co{s}^2\theta +si{n}^2\theta }{si{n}^2\theta .co{s}^2\theta }$

Also, we know that $co{s}^2\theta +si{n}^2\theta =1$

$\therefore \frac{co{s}^2\theta +si{n}^2\theta }{si{n}^2\theta .co{s}^2\theta }=se{c}^2\theta .cose{c}^2\theta$