MCQ
If $\sin\alpha+\sin\beta=\text{a}$ and $\cos\alpha-\cos\beta=\text{b}$ then $\tan\frac{\alpha-\beta}{2}=$
  • A
    $-\frac{\text{a}}{\text{b}}$
  • B
    $-\frac{\text{b}}{\text{a}}$
  • C
    $\sqrt{\text{a}^2+\text{b}^2}$
  • D
    None of these

Answer

  1. $-\frac{\text{b}}{\text{a}}$

Solution:

Given:

$\sin\alpha+\sin\beta=\text{a}$

$\Rightarrow2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha+\beta}{2}=\text{a}\ .....(1)$

Also,

$\cos\alpha+\cos\beta=\text{b}$

$\Rightarrow-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha+\beta}{2}=\text{b}\ .....(2)$

On dividing (1) by (2), we get

$\frac{-\cos\frac{\alpha+\beta}{2}}{\sin\frac{\alpha-\beta}{2}}=\frac{\text{a}}{\text{b}}$

$\Rightarrow\frac{-\sin\frac{\alpha+\beta}{2}}{\cos\frac{\alpha-\beta}{2}}=\frac{\text{b}}{\text{a}}$

$\Rightarrow\tan\frac{\alpha-\beta}{2}=-\frac{\text{b}}{\text{a}}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free