Question

Express $ta{n}^{-1}\left(\frac{cosx-sinx}{cosx+sinx}\right)$, for $0<x<\pi$ in the simplest form.

Answer 1

Let $y=ta{n}^{-1}\left(\frac{cosx-sinx}{cosx+sinx}\right)$

Then $tan\left(y\right)=\left(\frac{cosx-sinx}{cos+sinx}\right)$

We know that $sin\frac{\pi }{4}$ = $cos\frac{\pi }{4}$ = $\frac{1}{\sqrt{2}}$

Multiplying numerator and denominator with $\frac{1}{\sqrt{2}}$

Therefore, $tany=\frac{sin\frac{\pi }{4}cosx-cos\frac{\pi }{4}sinx}{cos\frac{\pi }{4}cosx+sin\frac{\pi }{4}sinx}$

Using the formulae, $sin(A-B)=sinAcosB-cosAsinB$ and $cos(A-B)=cosAcosB+sinASinB$,

$tany=\frac{sin(\frac{\pi }{4}-x)}{cos(\frac{\pi }{4}-x)}$ = $tan(\frac{\pi }{4}-x)$

Therefore principal value of $y=ta{n}^{-1}tan(\frac{\pi }{4}-x)$

Principal value of $ta{n}^{-1}$ must lie in $(-\frac{\pi }{2},\frac{\pi }{2})$.

Therefore $\frac{\pi }{4}-x$ must lie in $(-\frac{\pi }{2},\frac{\pi }{2})$.

$-\frac{\pi }{2}<\frac{\pi }{4}-x<\frac{\pi }{2}$

$-\frac{3\pi }{4}<-x<\frac{\pi }{4}$

$\frac{3\pi }{4}>x>-\frac{\pi }{4}$

It is given that $0<x<\pi$

Hence for $0<x<\frac{3\pi }{4}$

$y=\frac{\pi }{4}-x$