Question

$\frac{tan\theta }{(1+ta{n}^2\theta {)}^2}+\frac{cot\theta }{(1+co{t}^2\theta {)}^2}$ is equal to

• A. $2sin\theta .cos\theta$
• B. $cosec\theta .sec\theta$
• C. $sin\theta .cos\theta$
• D. $2cosec\theta .sin\theta$

Answer 1

The correct option is C $sin\theta .cos\theta$

$\frac{tan\theta }{(1+ta{n}^2\theta {)}^2}+\frac{cot\theta }{(1+co{t}^2\theta {)}^2}=\frac{tan\theta }{se{c}^4\theta }+\frac{cot\theta }{cose{c}^4\theta }$
$[\therefore se{c}^2\theta =1+ta{n}^2\theta ,cose{c}^2\theta =1+co{t}^2\theta ]$
$=sin\theta co{s}^3\theta +cos\theta si{n}^3\theta$
$=sin\theta cos\theta (co{s}^2\theta +si{n}^2\theta )$
$=sin\theta cos\theta$