Question

If $\sec \theta +\tan \theta =x$ then the value of $\sin \theta$ is

• A. $\frac{1-2x}{1+x^2}$
• B. $\frac{x^2-1}{x^2+1}$
• C. $\frac{1+x^2}{2(1-x^2)}$
• D. $\frac{x^2+1}{x^2-1}$

Answer 1

The correct option is B $\frac{x^2-1}{x^2+1}$

$\sec \theta +\tan \theta =x$
$\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=x$
$1+\sin \theta =x\cos \theta$
Squaring both sides,
$1+{\sin }^2\theta +2\sin \theta =x^2{\cos }^2\theta$
$1+si{n}^2\theta +2\sin \theta =x^2(1-{\sin }^2\theta )$
$(1+x^2){\sin }^2\theta +2\sin \theta +(1-x^2)=0$
Thus, $\sin \theta =\frac{-2\pm \sqrt{4-4(1-x^2)(1+x^2)}}{2(1+x^2)}$
$\sin \theta =\frac{-2\pm \sqrt{(4-4+4x^4)}}{2(1+x^2)}$
$\sin \theta =\frac{-2\pm 2x^2}{2(1+x^2)}$
$\sin \theta =-1,\frac{x^2-1}{x^2+1}$
Hence, from the options,
$\sin \theta =\frac{x^2-1}{x^2+1}$