Question

A circle centered at $O$ has radius $1$ and contains points $A$. Segment $AB$ is tangent to the circle at $A$ and $\angle AOB=\theta$. If point $C$ lies on $OA$, and $BC$ bisects the $\angle ABO$, then $OC$ equals

• A. $\sec \theta (\sec \theta -\tan \theta )$
• B. $\frac{{\cos }^2\theta }{1+\sin \theta }$
• C. $\frac{1}{1+\sin \theta }$
• D. $\frac{1-\sin \theta }{{\cos }^2\theta }$

Answer 1

Answer: (A), (C), (D)

$\sec \theta (\sec \theta -\tan \theta )$
$\frac{1}{1+\sin \theta }$
$\frac{1-\sin \theta }{{\cos }^2\theta }$

By sine rule in $△ABC,$

$\frac{\sin \left(\frac{90°-\theta }{2}\right)}{x}=\frac{\sin 90°}{BC}--------\left(1\right)$

By sine rule in $△OCB,$

$\frac{\sin \left(\frac{90°-\theta }{2}\right)}{1-x}=\frac{\sin 90°}{BC}--------\left(2\right)$

$\frac{eq\left(1\right)}{eq\left(2\right)}=\frac{1-x}{x}=\frac{1}{\sin \theta }$

$x=\sin \theta -x\sin \theta$

$x=\frac{\sin \theta }{1+\sin \theta }$

And $OC$, equals $\frac{1}{(1+\sin \theta )}$

$\left(A\right)\frac{1}{\cos \theta }(\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta })\Rightarrow \frac{1-\sin \theta }{\cos {}^2\theta }\Rightarrow \frac{1-\sin \theta }{1-\sin {}^2\theta }\Rightarrow \frac{1}{(1+\sin \theta )}\left[Correct\right]$

$\left(B\right)\frac{\cos {}^2\theta }{1+\sin \theta }\Rightarrow \frac{1-\sin {}^2\theta }{1+\sin \theta }\Rightarrow (1-\sin \theta )\left[Wrong\right]$

$\left(C\right)\frac{1}{1+\sin \theta }\left[Correct\right]$

$\left(D\right)\frac{1-\sin \theta }{\cos {}^2\theta }=\frac{1-\sin \theta }{1-\sin {}^2\theta }\Rightarrow \frac{1-\sin \theta }{(1+\sin \theta \left)\right(1-\sin \theta )}=\frac{1}{1+\sin \theta }\left[Correct\right]$

Hence, $\sec \theta \left(\sec \theta -\tan \theta \right)$ , $\frac{1}{1+\sin \theta }$ and $\frac{1-\sin \theta }{\cos {}^2\theta }$ are the correct answers.