Question

$1-\frac{si{n}^2\theta }{1+cot\theta }-\frac{co{s}^2\theta }{1+tan\theta }$

Answer 1

$LHS=1-\frac{si{n}^2\theta }{1+cot\theta }-\frac{co{s}^2\theta }{1+tan\theta }$

$=1-\frac{si{n}^2\theta }{1+\frac{cos\theta }{sin\theta }}-\frac{co{s}^2\theta }{1+\frac{sin\theta }{cos\theta }}(\because cot\theta =\frac{cos\theta }{sin\theta },tan\theta =\frac{sin\theta }{cos\theta })$

$=1-\frac{si{n}^2\theta }{\frac{sin\theta +cos\theta }{sin\theta }}-\frac{co{s}^2\theta }{\frac{cos\theta +sin\theta }{cos\theta }}=1-\frac{si{n}^3\theta }{sin\theta +cos\theta }-\frac{co{s}^3\theta }{cos\theta +sin\theta }$

$=\frac{sin\theta +cos\theta -(si{n}^3+co{s}^3\theta )}{sin\theta +cos\theta }$

$=\frac{sin\theta +cos\theta -(sin\theta +cos\theta )(si{n}^2\theta +co{s}^2\theta -sin\theta cos\theta )}{sin\theta +cos\theta }$

[Using $a^3+b^3=(a+b)(a^2+b^2-ab)]$

$=\frac{(sin\theta +cos\theta )[1-(1-sin\theta cos\theta \left)\right]}{sin\theta +cos\theta }$ [Using $si{n}^2\theta +co{s}^2\theta =1]$

$=sin\theta cos\theta =RHS$ Hence proved.