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 $\frac{1-p^2}{1+p^2}$

$\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }=\frac{1-\sin \theta }{\cos \theta }=p{\left(\frac{1-\sin \theta }{\cos \theta }\right)}^2=p^2\Rightarrow \frac{1-2\sin \theta +{\sin }^2\theta }{{\cos }^2\theta }=p^2\Rightarrow {\sin }^2\theta -2\sin \theta +1=p^2(1-{\sin }^2\theta )\Rightarrow (1+p^2){\sin }^2\theta -2\sin \theta +(1-p^2)=0\Rightarrow \sin \theta =\frac{2\pm \sqrt{4-4(1-p^4)}}{2(1+p^2)}\Rightarrow \sin \theta =\frac{1\pm p^2}{1+p^2}\Rightarrow \sin \theta =1and\sin \theta =\frac{1-p^2}{1+p^2}$