Question

$\frac{ta{n}^2\theta }{1+ta{n}^2\theta }$ + $\frac{co{t}^2\theta }{1+co{t}^2\theta }$ =

• A. 0
• B. -1
• C. 1
• D. 2

Answer 1

The correct option is C

1

$\frac{ta{n}^2\theta }{se{c}^2\theta }$ + $\frac{co{t}^2\theta }{cose{c}^2\theta }$

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