Question

$If\csc \theta -\cot \theta =k,thenprovethat\cos \theta =\frac{k^2-1}{k^2+1}$

Answer 1

$Given:\csc \theta +\cot \theta =kToprove:\cos \theta =\frac{k^2-1}{k^2+1}Proof:\csc \theta +\cot \theta =k\Rightarrow \frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta }=k\Rightarrow \frac{1+\cos \theta }{\sin \theta }=kNow,\frac{k^2-1}{k^2+1}=\frac{{\left(\frac{1+\cos \theta }{\sin \theta }\right)}^2-1}{{\left(\frac{1+\cos \theta }{\sin \theta }\right)}^2+1}=\frac{2{\cos }^2\theta +2\cos \theta }{2+2\cos \theta }=\cos \theta$