Question

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

Answer 1

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

$=\frac{\frac{{\sin }^3\theta }{{\cos }^3\theta }}{\frac{1}{{\cos }^2\theta }}+\frac{\frac{{\cos }^3\theta }{{\sin }^3\theta }}{\frac{1}{{\sin }^2\theta }}$

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

$=\frac{{\sin }^4\theta +{\cos }^4\theta }{\sin \theta \cos \theta }$

$=\frac{({\sin }^2\theta +{\cos }^2\theta {)}^2-2{\sin }^2\theta {\cos }^2\theta }{\sin \theta \cos \theta }$

$=\frac{1-2{\sin }^2\theta {\cos }^2\theta }{\sin \theta \cos \theta }$ ...............$\because [sin²\theta +cos²\theta =1]$

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

$=RHS$

Hence proved