Question

If $\sec \varphi +\tan \varphi =x,$ prove that $\sin \varphi =\frac{x^2-1}{x^2+1}$.

Answer 1

$\sec \varphi +\tan \varphi =x$,

To prove : $\sin \varphi =\frac{x^2-1}{x^2+1}$

$LHS$

$\sec \varphi +\tan \varphi =\frac{1}{\cos \varphi }+\frac{\sin \varphi }{\cos \varphi }=\frac{1+\sin \varphi }{\cos \varphi }$

$\frac{1+\sin \varphi }{\cos \varphi }=x$

Squaring both sides:

$\frac{(1+\sin \varphi {)}^2}{{\cos }^2\varphi }=x^2$

$\Rightarrow \frac{(1+\sin \varphi {)}^2}{1-{\sin }^2\varphi }=x^2(\therefore {\cos }^2\varphi =1-{\sin }^2\varphi )$

$\Rightarrow \frac{(1+\sin \varphi {)}^2}{(1+\sin \varphi )(1-\sin \varphi )}=x^2(\therefore a^2-b^2=(a+b\left)\right(a-b\left)\right)$

$\Rightarrow \frac{1+\sin \varphi }{1-\sin \varphi }=x^2$

$\Rightarrow 1+\sin \varphi =x^2-x^2\sin \varphi$

$\Rightarrow \sin \varphi (1+x^2)=x^2-1$

$\Rightarrow \sin \varphi =\frac{x^2-1}{x^2+1}$ Hence proved