Question

If $\sec \Theta -\tan \Theta =p,$ then the value of $\sin \Theta$ is

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

Answer 1

The correct option is D $\frac{1-p^2}{1+p^2}$

Consider the given equation,

$\sec \theta -\tan \theta =p$ …….(1)

Since equation (1),

$\sec \theta -\tan \theta =p$

$\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }=p$

$\frac{1-\sin \theta }{\cos \theta }=p$

$\frac{\cos \theta }{1-\sin \theta }=\frac{1}{p}$

$\frac{\cos \theta }{1-\sin \theta }\times \frac{1+\sin \theta }{1+\sin \theta }=\frac{1}{p}$

$\frac{\cos \theta (1+\sin \theta )}{1-{\sin }^2\theta }=\frac{1}{p}$

$\frac{1+\sin \theta }{\cos \theta }=\frac{1}{p}$

$\frac{\cos \theta }{1-\sin \theta }=p$

$\sec \theta +\tan \theta =\frac{1}{p}$ …….(2)

Since, equation (1)and (2),

$2\sec \theta =p+\frac{1}{p}$

$\sec \theta =\frac{p^2+1}{2p}$

$\cos \theta =\frac{2p}{p^2+1}$

Now,

$\sin \theta =\sqrt{1-{\cos }^2\theta }$

$=\sqrt{1-{\left(\frac{2p}{p^2+1}\right)}^2}$

$=\sqrt{\frac{{(p^2+1)}^2-4p^2}{{(p^2+1)}^2}}=\sqrt{\frac{{(p^2-1)}^2}{{(p^2+1)}^2}}$

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

Hence, this is the answer.