Question

If $y\cos \theta -1=\sin \theta$, then $\sin \theta$ is

• A. $\sin \theta =\frac{y^2-1}{y^2+1}$
• B. $\sin \theta =\frac{1}{y^2+1}$
• C. $\sin \theta =\frac{y^2}{y^2+1}$
• D. $\sin \theta =\frac{y-1}{y^2+1}$

Answer 1

The correct option is A $\sin \theta =\frac{y^2-1}{y^2+1}$

Given $y\cos \theta -1=\sin \theta$, so
$\Rightarrow y=\frac{1+\sin \theta }{\cos \theta }\Rightarrow y^2=\frac{(1+\sin \theta {)}^2}{{\cos }^2\theta }\Rightarrow y^2=\frac{(1+\sin \theta {)}^2}{1-{\sin }^2\theta }\Rightarrow y^2=\frac{1+\sin \theta }{1-\sin \theta }\Rightarrow y^2-y^2\sin \theta =1+\sin \theta \Rightarrow (y^2+1)\sin \theta =y^2-1\therefore \sin \theta =\frac{y^2-1}{y^2+1}$