Question

Prove that

$\frac{tan({90}^{}-\theta )}{cosec\theta +1}+\frac{cosec\theta +1}{cot\theta }=2sec\theta$

Answer 1

$LHS=\frac{\tan (90-\theta )}{\csc \theta +1}+\frac{\csc \theta +1}{\cot \theta }$

$=\frac{\cot \theta }{\csc \theta +1}+\frac{\csc \theta +1}{\cot \theta }$

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

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

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

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

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

$=\frac{2\csc \theta }{\cot \theta }$

$=2\csc \theta \tan \theta$

$=2\times \frac{1}{\sin \theta }\times \frac{\sin \theta }{\cos \theta }$

$=2\sec \theta =RHS$

Hence proved.