Question

If $cosec\theta +cot\theta =k$ then prove that $cos\theta =\frac{k^2-1}{k^2+1}$

Answer 1

$cosec\theta +\cot \theta =k$

$\therefore \frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta }=k$

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

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

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

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

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

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

$=\cos \theta$

$\cos \theta =\frac{k^2-1}{k^2+1}$

Hence proved