<s>[INST] if $\frac{\sin(2a-b)}{\sin b}=1+x$ prove that... 
if $\frac{\sin(2a-b)}{\sin b}=1+x$ prove that $$\frac{\cos b}{ \cos a}=1+\frac{1}{2}x\tan^2 a$$ approximately .(Take $x$ to be very small)


Attempt by given eq. $$\sin 2a \cos b-\sin b \cos 2a=\sin b +x\sin b$$
or $$2\sin a \cos a \cos b -\sin b (1-2\sin^2 a)=\sin b +x\sin b$$ .
or $$\frac{\cos b}{\cos a}=\frac{1}{2\sin b}\sec^2 a \left(\sin b (1-2\sin^2 a)+\sin b +x\sin b \right)$$ .
How do i proceed?
 [/INST] $\begin{align}\frac{\cos b}{\cos a} &= \frac{\sec^2a}{2}\left[1-2\sin^2a+(1+x)\right] = \frac{\sec^2a}{2} - \tan^2a +\frac{1+x}{2}\sec^2a 
\\&= \left(1+\frac x2\right)\sec^2a -\tan^2a = \left(1+\frac x2\right)(1+\tan^2a)-\tan^2a = \left(1+\frac x2\right)+\frac x2\tan^2a \\&\approx1+\frac x2\tan^2a\end{align} $
As $x$ is very small, you may take $1+\frac x2 \approx 1$
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