Question

If $0<\beta <\alpha <\frac{\pi }{2},\cos (\alpha +\beta )=\frac{3}{5}and\cos (\alpha -\beta )=\frac{4}{5}$, then $\sin 2\alpha$ is equal to.

• A. $0$
• B. $1$
• C. $2$
• D. $Noneofthese$

Answer 1

The correct option is B

$1$

Step 1. Framing equations

Given, $\cos (\alpha +\beta )=\frac{3}{5}\&\cos (\alpha -\beta )=\frac{4}{5}$

Now,

$\begin{array}{rcl}\cos (\alpha +\beta )& =& \frac{3}{5}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}(\alpha +\beta )& =& {\cos }^{-1}\left(\frac{3}{5}\right)\end{array}$…..(i)

$\begin{array}{rcl}\cos (\alpha -\beta )& =& \frac{4}{5}\end{array}$

$\Rightarrow$$\begin{array}{rcl}sin(\alpha -\beta )& =& \frac{3}{5}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}(\alpha -\beta )& =& si{n}^{-1}\left(\frac{3}{5}\right)\end{array}$…..(ii)

Step 2. Add equation (i) & (ii), we get

$\begin{array}{rcl}\alpha +\beta +\alpha -\beta & =& {\cos }^{-1}\left(\frac{3}{5}\right)+si{n}^{-1}\left(\frac{3}{5}\right)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}2\alpha & =& \frac{\pi }{2}\end{array}$

$\begin{array}{rcl}\therefore sin2\alpha & =& \sin \frac{\pi }{2}\\ & =& \mathbf{1}\end{array}$

Hence, correct option is $\left(B\right)$