Question

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

Answer 1

Let $LHS=\frac{{\cot }^3\theta }{1+{\cot }^2\theta }+\frac{{\tan }^3\theta }{1+{\tan }^2\theta }$

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

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

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

$=\frac{(1-{\sin }^2\theta )\cos \theta }{\sin \theta }+\frac{(1-{\cos }^2\theta )\sin \theta }{\cos \theta }$

$=\frac{\cos \theta }{\sin \theta }-\sin \theta \cos \theta +\frac{\sin \theta }{\cos \theta }-\sin \theta .\cos \theta$

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

$=\frac{1}{\sin \theta .\cos \theta }-2\sin \theta \cos \theta$

$=\csc \theta .\sec \theta -2\sin \theta \cos \theta$

$=RHS$ Hence proved