Question

Assertion :If $\alpha$ and $\beta$ are two distinct solution of the equations $a\cos x+b\sin x=c$ then $\tan \left(\frac{\alpha +\beta }{2}\right)$ is independent of c. Reason: Solution of $a\cos x+b\sin x=c$ is possible if $-\sqrt{a^2+b^2}\le c\le \sqrt{a^2+b^2}$

• A. Statement-1 is false, statement-2 is true
• B. Statement-1 is true, statement-2 is true,statement-2 is a correct explanation for statement-1
• C. Statement-1 is true, statement-2 is true,statement-2 is not a correct explanation for statement-1
• D. Statement-1 is true, statement-2 is false

Answer 1

The correct option is C Statement-1 is true, statement-2 is true,statement-2 is not a correct explanation for statement-1

$a\cos x+b\sin x=c$
$\Rightarrow a\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)+\frac{2b\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=c$
$\Rightarrow (a+c){\tan }^2\frac{x}{2}-2b\tan \frac{x}{2}+(c-a)=0$
The equation has roots $\tan \frac{\alpha }{2}$ and $\tan \frac{\beta }{2}$
$\therefore \tan \frac{\alpha }{2}\tan \frac{\beta }{2}=\frac{c-a}{a+c}$
$\tan \left(\frac{\alpha +\beta }{2}\right)=\frac{\tan \frac{\alpha }{2}\tan \frac{\beta }{2}}{1-\tan \frac{\alpha }{2}\tan \frac{\beta }{2}}=\frac{\frac{2b}{a+c}}{1-\frac{c-a}{a+c}}=\frac{b}{a}$