Question

If $cosec\theta +cot\theta =m$, show that $\frac{m^2-1}{m^2+1}=cos\theta$

Answer 1

Given : $cosec\theta +\cot \theta =m$

$=(1+\cos \theta )/\sin \theta$

$m^2-1\Rightarrow (1+{\cos }^2\theta +2\cos \theta -{\sin }^2\theta$

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

$\Rightarrow m^2-1=\frac{\cos \theta {\cos }^2\left(\theta /2\right)}{{\sin }^2\theta }$

$m^2+1\Rightarrow 1+{\cos }^2\theta +2\cos \theta +\frac{{\sin }^2\theta }{{\sin }^2\theta }$

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

$\Rightarrow 4{\cos }^2\left(\frac{\theta }{2}\right)/{\sin }^2\theta$

$\frac{(m^2-1)}{m^2+1}\Rightarrow \left[4\cos \theta {\cos }^2\right(\theta /2\left)/{\sin }^2\theta \right]/4{\cos }^2\left(\theta /2\right)/{\sin }^2\theta ]=\cos \theta$

Hence proved