prove the following

**IIT 2010** · September 13th 2009, 09:31 AM

prove
tan(x-(a/2)) = [(k-1)tan(a/2)]/(k+1)
given that
sinx=k sin(a-x)
my ans is slightly different
on rhs denominator i get 1+ktan^2(a/2)

**Grandad** · September 13th 2009, 10:41 PM

Hello IIT 2010

Originally Posted by IIT 2010

prove
tan(x-(a/2)) = [(k-1)tan(a/2)]/(k+1)
given that
sinx=k sin(a-x)
my ans is slightly different
on rhs denominator i get 1+ktan^2(a/2)

$\sin x = k\sin(a-x)$

$\Rightarrow \sin x = k(\sin a\cos x - \cos a\sin x)$

$\Rightarrow \sin x(1+k\cos a) = k\sin a\cos x$

$\Rightarrow \tan x = \frac{k\sin a}{1+k\cos a}$

$= \frac{\dfrac{2kt}{1+t^2}}{1+\dfrac{k(1-t^2)}{1+t^2}}$ , where $t =\tan \tfrac12a$

$=\frac{2kt}{1+t^2+k(1-t^2)}$

$\Rightarrow \tan(x-\tfrac12a)=\frac{\tan x -\tan\tfrac12a}{1+\tan x \tan\tfrac12a}$

$=\frac{\dfrac{2kt}{1+t^2+k(1-t^2)} -t}{1+\dfrac{2kt^2}{1+t^2+k(1-t^2)}}$

$=\frac{2kt-t\Big(1+t^2+k(1-t^2)\Big)}{1+t^2+k(1-t^2)+2kt^2}$

$=\frac{t(2k-1-t^2-k+kt^2)}{1+t^2+k-kt^2+2kt^2}$

$=\frac{t\Big((k-1)+t^2(k-1)\Big)}{(k+1)+t^2(k+1)}$

$=\frac{t(k-1)(1+t^2)}{(k+1)(1+t^2)}$

$=\frac{(k-1)\tan\tfrac12a}{(k+1)}$

Grandad

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