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If $$\frac{m+1}{m-1}=\frac{\cos(\alpha-\beta)}{\sin(\alpha+\beta)}$$, prove that : $$m=\tan(π/4 +\alpha).\tan(π/4 +\beta)$$.

My attempts/

Here .. $$\frac{m+1}{m-1}=\frac{\cos(\alpha-\beta)}{\sin(\alpha+\beta)}$$ Applying componendo and dividendo, I got; $$m=\frac{\cos(\alpha-\beta)+\sin(\alpha+\beta)}{\cos(\alpha-\beta)-\sin(\alpha+\beta)}$$.

Now, how should I move on?

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Hint:

Set $\alpha+\dfrac\pi4=A\iff\alpha=A-\dfrac\pi4$

and $\beta+\dfrac\pi4\iff\beta=B-\dfrac\pi4$

Now apply Componendo and dividendo

Alternatively,

write $\alpha-\beta=\alpha+\dfrac\pi4-\left(\beta+\dfrac\pi4\right)$

$\alpha+\beta=\alpha+\dfrac\pi4+\beta+\dfrac\pi4-\dfrac\pi2$

and use $\sin\left(A-\dfrac\pi2\right)=-\sin\left(\dfrac\pi2-A\right)=-\cos A$

Now apply Componendo and dividendo

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  • $\begingroup$ could you expand a bit please. $\endgroup$ – Iaamuser user Feb 7 '16 at 13:52
  • $\begingroup$ @Iaamuseruser, Please find the updated version $\endgroup$ – lab bhattacharjee Feb 7 '16 at 14:36
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Notice, the following trig identities $\cos A-\cos B=2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{B-A}{2}\right)$ & $\cos A+\cos B=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$, now one should get

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

$$=\frac{\sin\left(\frac{\pi}{4}+\alpha\right)\sin\left(\frac{\pi}{4}+\beta\right)}{\cos\left(\frac{\pi}{4}+\alpha\right)\cos\left(\frac{\pi}{4}+\beta\right)}$$ $$=\tan\left(\frac{\pi}{4}+\alpha\right)\tan\left(\frac{\pi}{4}+\beta\right)$$

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