Question

lf ${\theta }_1,{\theta }_2,{\theta }_3,\dots ..{\theta }_n$ are in $A.P.$, then $\frac{\sin {\theta }_1+{\sin }_2+....+\sin {\theta }_n}{\cos {\theta }_1+\cos {\theta }_2+.+\cos {\theta }_n}=$

• A. $0$
• B. $\tan ({\theta }_1+{\theta }_n)$
• C. $\tan \left(\frac{{\theta }_1+{\theta }_n}{2}\right)$
• D. $\tan \left(\frac{{\theta }_n-{\theta }_1}{2}\right)$

Answer 1

The correct option is C $\tan \left(\frac{{\theta }_1+{\theta }_n}{2}\right)$

${\theta }_1.......{\theta }_n$ are in A.P

$\therefore \Rightarrow \frac{{\theta }_1+{\theta }_n}{2}=\frac{{\theta }_2+{\theta }_{n-1}}{2}=\frac{{\theta }_3+{\theta }_{n-2}}{2}$

${\theta }_n={\theta }_1+(n-1)\underset{d}{\underset{}{({\theta }_1-{\theta }_2)}}$

${\theta }_n+\theta 1=2{\theta }_1+(n-1)d$

$\Rightarrow {\theta }_1+{\theta }_n={\theta }_2+{\theta }_{n-1}={\theta }_3+{\theta }_{n-2}=\cdots =2{\theta }_1+(n-1)d$

$\sin {\theta }_1+\sin {\theta }_2.....\sin {\theta }_n=\sin {\theta }_1+\sin {\theta }_n+\sin {\theta }_2+\sin {\theta }_{n-1}$

$=2\sin \left(\frac{{\theta }_1+{\theta }_2}{2}\right)\cos \left(\frac{{\theta }_1-{\theta }_n}{2}\right)+2\sin \left(\frac{{\theta }_n+{\theta }_{n-1}}{2}\right)\cos \left(\frac{{\theta }_{n-1}-{\theta }_2}{2}\right)$.... in case of $\cos \theta$

$\cos {\theta }_1+\cos {\theta }_2-......\cos {\theta }_n$

$=\cos {\theta }_1+\cos {\theta }_n+\cos {\theta }_2+\cos {\theta }_{n-1}....$

$=2\cos \left(\frac{{\theta }_1+{\theta }_n}{2}\right)\cos \frac{{\theta }_n-{\theta }_1}{2}+2\cos ({\theta }_2+{\theta }_{n-1})+2\cos \left(\frac{{\theta }_2+{\theta }_{n-1}}{2}\right)\cos \left(\frac{\theta }{2}\right)\cos \left(\frac{{\theta }_{n-1}-{\theta }_2}{2}\right)$

$\therefore \frac{\sin {\theta }_1+.....\sin {\theta }_n}{\cos {\theta }_1+\cos {\theta }_n}=\frac{2\sin \left(\frac{{\theta }_1+{\theta }_n}{2}\right)\cos \left(\frac{{\theta }_1-{\theta }_n}{2}\right)+2\sin \left(\frac{{\theta }_1+{\theta }_{n-1}}{2}\right)cos\left(\frac{{\theta }_2-{\theta }_{n-1}}{2}\right)}{2\cos \left(\frac{{\theta }_1+{\theta }_n}{2}\right)\cos \frac{{\theta }_n-{\theta }_1}{2}+2\cos \left(\frac{{\theta }_2+{\theta }_{n-1}}{2}\right)cos\left(\frac{{\theta }_2-{\theta }_{n-1}}{2}\right)}$

$=\frac{\sin \left(\frac{{\theta }_1+{\theta }_n}{2}\right)}{\cos \left(\frac{{\theta }_1+{\theta }_n}{2}\right)}$

$=\tan \left(\frac{{\theta }_1+{\theta }_n}{2}\right)$