Question

Prove the following: $\tan \theta +\tan (90°-\theta )=sec\theta ·sec(90°-\theta )$

Answer 1

To prove $\tan \theta +\tan (90°-\theta )=sec\theta ·sec(90°-\theta )$

Consider LHS:

$\begin{array}{rcl}\mathrm{LHS}& =& \tan \theta +\tan (90°-\theta )\\ & =& \tan \theta +cot\theta \left[\because \tan (90°-\theta )=\cot \theta \right]\\ & =& \frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }\left[\because \tan \theta =\frac{\sin \theta }{\cos \theta }\mathrm{and}\cot \theta =\frac{\cos \theta }{\sin \theta }\right]\\ & =& \frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta ·\cos \theta }\\ & =& \frac{1}{\sin \theta ·\cos \theta }\left[\because {\sin }^2\theta +{\cos }^2\theta =1\right]\\ & =& \sec \theta ·\mathrm{cosec}\theta \left[\frac{1}{\sin \theta }=\mathrm{cosec}\theta \mathrm{and}\frac{1}{\cos \theta }=\sec \theta \right]\\ & =& \sec \theta ·\sec (90°-\theta )\left[\because \mathrm{cosec}\theta =\sec (90°-\theta )\right]\\ & =& \mathrm{RHS}\end{array}$

Thus,$\mathrm{LHS}=\mathrm{RHS}$

Hence,$\tan \theta +\tan (90°-\theta )=sec\theta ·sec(90°-\theta )$ is proved