Question

$\begin{array}{c}if\sec \theta +\tan \theta =h,provethat\\ \frac{h^2-1}{h^{2}1}=\sin \theta \end{array}$

Answer 1

Given,

$\sec \theta +\tan \theta =h$

To prove:- $\frac{h^2-1}{h^2+1}=\sin \theta$

Proof:-

$\sec \theta +\tan \theta =h$

$\Rightarrow \frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=h$

$\Rightarrow \frac{1+\sin \theta }{\cos \theta }=h$

Squaring both the sides,

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

$\Rightarrow \frac{(1+\sin \theta {)}^2}{(1-{\sin }^2\theta )}=h^2$

$\Rightarrow \frac{(1+\sin \theta {)}^2}{(1-\sin \theta )(1+\sin \theta )}=h^2$

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

Now using componendo & dividendo, we get

$\Rightarrow \frac{1+\sin \theta -(1-\sin \theta )}{1+\sin \theta +1-\sin \theta }=\frac{h^2-1}{h^2+1}$

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

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