Question

Show that : $\frac{{tan}^3\theta }{1+{tan}^2\theta }+\frac{{cot}^3\theta }{1+{cot}^2\theta }=sec\theta cosec\theta -2sin\theta cos\theta$

Answer 1

$\frac{ta{n}^3\theta }{1+ta{n}^2\theta }+\frac{co{t}^3\theta }{1+co{t}^2\theta }=sec\theta cosec\theta -2sin\theta cos\theta$

LHS

$\Rightarrow \frac{ta{n}^3\theta }{se{c}^2\theta }+\frac{\theta \left(co{s}^3\theta /si{n}^3\theta \right)}{(co{s}^2\theta +se{c}^2\theta )}\times si{n}^2\theta$

$\Rightarrow \frac{si{n}^3\theta }{co{s}^3\theta }co{s}^2\theta +\frac{co{s}^3\theta }{sin\theta }=[\frac{si{n}^3}{cos\theta }+\frac{co{s}^3\theta }{sin\theta }]$

$\Rightarrow \frac{si{n}^4\theta +co{s}^4\theta }{sin\theta cos\theta }\Rightarrow \frac{(si{n}^2+co{s}^2\theta {)}^2-2si{n}^2\theta co{s}^2\theta }{sin\theta cos\theta }$

$\Rightarrow \frac{\left(1\right)-2si{n}^2\theta co{s}^2\theta }{sin\theta cos\theta }\Rightarrow \left(\frac{1}{sin\theta }\right)\left(\frac{1}{cos\theta }\right)-2\frac{si{n}^2\theta co{s}^2\theta }{sin\theta cos\theta }$

$\Rightarrow sec\theta cosec\theta -2sin\theta cos\theta =RHS$

$\therefore LHS=RHS$