Question

If $\frac{\cos x}{\cos y}=\frac{a}{b}$, show that,
$a\tan x+b\tan y=(a+b)\tan \frac{x+y}{2}$.

Answer 1

Given that,

$\frac{\cos x}{\cos y}=\frac{a}{b}$ ……. (1)

Then, prove that,

$a\tan x+b\tan y=(a+b)\tan \frac{x+y}{2}$

Or

$\frac{a\tan x+b\tan y}{(a+b)}=\tan \frac{x+y}{2}$

Proof:- By equation (1)

$b\cos x=a\cos y$

$\frac{b}{a}=\frac{\cos y}{\cos x}$

$(\frac{b}{a}+1)=(\frac{\cos y}{\cos x}+1)$

$\frac{a+b}{a}=\frac{\mathrm{cosx}+cosy}{\cos x}$

$\frac{a}{a+b}=\frac{\cos x}{\mathrm{cosx}+cosy}$ …… (2)

Similarly,

$\frac{b}{a+b}=\frac{\cos y}{\mathrm{cosx}+cosy}$ …… (3)

L.H.S.

$\frac{a\tan x+b\tan y}{(a+b)}$

$=\frac{a}{a+b}\tan x+\frac{b}{a+b}\tan y$

Put the value by equation (2) and (3) and we get,

$=\frac{\cos x}{\cos x+\cos y}\tan x+\frac{\mathrm{cosy}}{\cos x+\cos y}\tan y$

$=\frac{\cos x\tan x+\cos y\tan y}{\cos x+\cos y}$

$=\frac{\cos x\frac{\sin x}{\cos x}+\cos y\frac{\sin y}{\cos y}}{\cos x+\cos y}$

$=\frac{\sin x+\sin y}{\cos x+\cos y}$

$=\frac{2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)}{2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)}$

$=\tan \left(\frac{x+y}{2}\right)$

R.H.S.

Hence proved.