Question

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

Answer 1

$LHS=\frac{ta{n}^3\theta }{1+ta{n}^2\theta }+\frac{co{t}^3\theta }{1+co{t}^2\theta }$

$=\frac{si{n}^2\theta }{co{s}^3(1+\frac{si{n}^2\theta }{co{s}^2\theta })}+\frac{co{s}^3\theta }{si{n}^3(1+\frac{co{s}^2\theta }{si{n}^2\theta })}=(\because tan\theta =\frac{sin\theta }{cos\theta }andcot\theta =\frac{cos\theta }{sin\theta })$

$=\frac{si{n}^3\theta co{s}^2\theta }{co{s}^3\theta (co{s}^2\theta +si{n}^2\theta )}+\frac{co{s}^3\theta si{n}^2\theta }{si{n}^3\theta (si{n}^2\theta +co{s}^2\theta )}$

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

$=\frac{si{n}^4\theta +co{s}^4\theta }{sin\theta cos\theta }$

$=\frac{\left(si{n}^2\theta {)}^2\right(co{s}^2\theta {)}^2+2si{n}^2\theta co{s}^2\theta -2si{n}^2\theta co{s}^2\theta }{sin\theta cos\theta }$ (adding and subtracting $2si{n}^2\theta co{s}^2\theta )$

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

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

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

Hence proved.