Question

If $\alpha ,\beta$ are differnt values of x satisfying $acosx+bsinx=c$, then $tan\left(\frac{\alpha +\beta }{2}\right)=$

• A. $a+b$
• B. $a-b$
• C. $\frac{b}{a}$
• D. $\frac{a}{b}$

Answer 1

The correct option is C $\frac{b}{a}$

$a\cos x+b\sin x=c$

$\Rightarrow a\cos x=c-b\sin x$

squaring both sides we get

$\Rightarrow a^2{\cos }^{2}x={(c-b\sin x)}^2$

$\Rightarrow a^2{\cos }^{2}x=c^2+b^2{\sin }^{2}x-2bc\sin x$

$\Rightarrow a^2(1-{\sin }^{2}x)=c^2+b^2{\sin }^{2}x-2bc\sin x$

$\Rightarrow a^2-a^2{\sin }^{2}x=c^2+b^2{\sin }^{2}x-2bc\sin x$

$\Rightarrow -a^2{\sin }^{2}x-b^2{\sin }^{2}x+2bc\sin x+(a^2-c^2)=0$

$\Rightarrow -(a^2+b^2){\sin }^{2}x+2bc\sin x+(a^2-c^2)=0$

$\Rightarrow (a^2+b^2){\sin }^{2}x-2bc\sin x+(c^2-a^2)=0$

Let $\sin \alpha$ and $\sin \beta$ be the roots of the equation then

Sum of the roots $=\sin \alpha +\sin \beta =\frac{2bc}{a^2+b^2}$ .....$\left(1\right)$

$a\cos x+b\sin x=c$

$\Rightarrow b\sin x=c-a\cos x$

squaring both sides we get

$\Rightarrow b^2{\sin }^{2}x={(c-a\cos x)}^2$

$\Rightarrow b^2{\sin }^{2}x=c^2+a^2{\cos }^{2}x-2ca\cos x$

$\Rightarrow b^2(1-{\cos }^{2}x)=c^2+a^2{\cos }^{2}x-2ca\cos x$

$\Rightarrow b^2-b^2{\cos }^{2}x=c^2+a^2{\cos }^{2}x-2ca\cos x$

$\Rightarrow -b^2{\cos }^{2}x-a^2{\cos }^{2}x+2ac\cos x+(b^2-c^2)=0$$\Rightarrow -(a^2+b^2){\sin }^{2}x+2ac\cos x+(b^2-c^2)=0$

$\Rightarrow (a^2+b^2){\cos }^{2}x-2ac\cos x+(c^2-b^2)=0$

Let $\cos \alpha$ and $\cos \beta$ be the roots of the equation then

Sum of the roots $=\cos \alpha +\cos \beta =\frac{2ac}{a^2+b^2}$ .....$\left(2\right)$

Eqn$\left(1\right)÷$Eqn$\left(2\right)$ we get

$\frac{\sin \alpha +\sin \beta }{\cos \alpha +\cos \beta }=\frac{\frac{2bc}{a^2+b^2}}{\frac{2ac}{a^2+b^2}}$

$\frac{2\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)}{2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)}=\frac{b}{a}$

$\therefore \tan \left(\frac{\alpha +\beta }{2}\right)=\frac{b}{a}$