Question

$\alpha ,\beta .\gamma ,\delta$ are eecentric angles of concyclic points of hyperbola, then $\cos \frac{(\alpha -\beta )}{2}\sin \frac{(\gamma -\delta )}{2}+\sin \frac{(\alpha +\beta )}{2}\cos \frac{(\gamma -\delta )}{2}=$

• A. $1$
• B. $\cos \alpha +\beta$
• C. $0$
• D. $\sin \alpha +\beta$

Answer 1

Answer: (C)

$0$

$\cos \left(\frac{\alpha -\beta }{2}\right)\sin \left(\frac{\gamma -\delta }{2}\right)+\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\gamma -\delta }{2}\right)\Rightarrow \cos \left(\frac{\alpha -\beta }{2}\right)\sin \left(\frac{\gamma -\delta }{2}\right)=-\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\gamma -\delta }{2}\right)\frac{\cos \left(\frac{\alpha -\beta }{2}\right)}{\cos \left(\frac{\gamma -\delta }{2}\right)}=\frac{-\sin \left(\frac{\alpha +\beta }{2}\right)}{\sin \left(\frac{\gamma -\delta }{2}\right)}\cos \left(\frac{A-B}{2}\right)=\cos \frac{A}{2}\cos \frac{B}{2}+\sin \frac{A}{2}\sin \frac{B}{2}\sin \left(\frac{A+B}{2}\right)=\sin \frac{A}{2}\cos \frac{B}{2}+\sin \frac{B}{2}\cos \frac{A}{2}\sin \left(\frac{A-B}{2}\right)=\sin \frac{A}{2}\cos \frac{B}{2}-\sin \frac{B}{2}\cos \frac{A}{2}\frac{\cos \frac{\alpha }{2}\cos \frac{\beta }{2}+\sin \frac{\alpha }{2}\sin \frac{\beta }{2}}{\cos \frac{\gamma }{2}\cos \frac{\delta }{2}+\sin \frac{\gamma }{2}\sin \frac{\delta }{2}}=\frac{-[\sin \frac{\alpha }{2}\cos \frac{\beta }{2}+\sin \frac{\beta }{2}\cos \frac{\alpha }{2}]}{\sin \frac{\gamma }{2}\cos \frac{\delta }{2}-\sin \frac{\delta }{2}\cos \frac{\gamma }{2}}$

For $LHS$, divide numerator by $\left[\cos \frac{\alpha }{2}\cos \frac{\beta }{2}\right]$ and denominator by $\left[\cos \frac{\gamma }{2}\cos \frac{\delta }{2}\right]$

For $RHS$, divide numerator by $\left[\cos \frac{\alpha }{2}\cos \frac{\beta }{2}\right]$ and denominator by $\left[\cos \frac{\gamma }{2}\cos \frac{\delta }{2}\right]$

$\Rightarrow \frac{1+\tan \frac{\alpha }{2}\tan \frac{\beta }{2}}{1+\tan \frac{\gamma }{2}\tan \frac{\delta }{2}}=-\left[\frac{\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}}{\tan \frac{\gamma }{2}-\tan \frac{\delta }{2}}\right]\frac{\tan \frac{\gamma }{2}-\tan \frac{\delta }{2}}{1+\tan \frac{\gamma }{2}\tan \frac{\delta }{2}}=\frac{-[\tan \frac{\alpha }{2}+\tan \frac{\beta }{2}]}{1+\tan \frac{\alpha }{2}\tan \frac{\beta }{2}}\Rightarrow \tan [\frac{\gamma }{2}-\frac{\delta }{2}]=-\tan [\frac{\alpha }{2}+\frac{\beta }{2}]\Rightarrow \gamma -\delta =-\alpha -\beta \alpha +\beta +\gamma =0\therefore \delta =\alpha +\beta +\gamma \Rightarrow \delta =0$