Question

If $\csc \theta -\cot \theta =k$, then show that $\cos \theta =\frac{1-k^2}{1+k^2}$.

Answer 1

Given that, $\csc \theta -cot\theta =k$

$RHS$

$\frac{1-k^2}{1+k^2}$

$=\frac{1-{(\csc \theta -\cot \theta )}^2}{1+{(\csc \theta -\cot \theta )}^2}$

$=\frac{1-({\csc }^2\theta +{\cot }^2\theta -2\csc \theta \cot \theta )}{1+({\csc }^2\theta +{\cot }^2\theta -2\csc \theta \cot \theta )}$

$=\frac{1-{\csc }^2\theta -{\cot }^2\theta +2\csc \theta \cot \theta }{1+{\csc }^2\theta +{\cot }^2\theta -2\csc \theta \cot \theta }$

$=\frac{-({\csc }^2\theta -1)-{\cot }^2\theta +2\csc \theta \cot \theta }{(1+{\cot }^2\theta )+{\csc }^2\theta -2\csc \theta \cot \theta }\therefore {\csc }^2\theta -1={\cot }^2\theta$

$=\frac{-{\cot }^2\theta -{\cot }^2\theta +2\csc \theta \cot \theta }{{\csc }^2\theta +{\csc }^2\theta -2\csc \theta \cot \theta }\therefore 1+{\cot }^2\theta ={\csc }^2\theta$

$=\frac{-2{\cot }^2\theta +2\csc \theta \cot \theta }{2{\csc }^2\theta -2\csc \theta \cot \theta }$

$=\frac{-2\cot \theta (\cot \theta -\csc \theta )}{2\csc \theta (\csc \theta -\cot \theta )}$

$=\frac{\cot \theta }{\csc \theta }$

$=\frac{\frac{\cos \theta }{\sin \theta }}{\frac{1}{\sin \theta }}$

$=\cos \theta$

$LHS$

Hence proved.