Question

Write the simplest form of $$$$ $ta{n^{ - 1}}\left( {\dfrac{{cosx - sinx}}{{cosx + sinx}}} \right),0 < x < 2\pi$ .

Answer 1

Hint : To write the simplest form of the given trigonometric expression firstly convert the sinx and cosx in terms of tanx by taking cosx from numerator and denominator. Once it gets converted into the tanx just write tan45 in place of 1 such that we will get the whole term as tan of some angle.

Complete step-by-step answer :
Given:
$ta{n^{ - 1}}\left( {\dfrac{{cosx - sinx}}{{cosx + sinx}}} \right)$
Taking $cosx$ common from numerator and denominator
We get,
$= {\tan ^{ - 1}}\left( {\dfrac{{1 - \dfrac{{sinx}}{{cosx}}}}{{1 + \dfrac{{sinx}}{{cosx}}}}} \right)$
on writing $tanx$ in place of $\dfrac{{sinx}}{{cosx}}$
we get,
$= ta{n^{ - 1}}\left( {\dfrac{{1 - tanx}}{{1 + tanx}}} \right)$
now replacing 1 as $\;tan45$ $= {\tan ^{ - 1}}\left( {\dfrac{{tan45 - tan x}}{{1 - tan 45\times tan x}}} \right)$
$= {\tan ^{ - 1}}\left( {\dfrac{{tan45 - tan x}}{{1 - tan 45\times tan x}}} \right)$
now by using $tan\left( {A - B} \right)$ formula we can write
$= ta{n^{ - 1}}\left[ {tan(45 - x)} \right]$
As we can cancel out the tan from above result
We get,
$45 - x$
Hence the value of
∴ $ta{n^{ - 1}}\left( {\dfrac{{cosx - sinx}}{{cosx + sinx}}} \right) = \dfrac{\pi }{4} - x$
Or we can say $\dfrac{\pi }{4} - x$ is the simplest form of the given inverse trigonometric function.
So, the correct answer is “$\dfrac{\pi }{4} - x$ ”.

Note : Here we have taken $cosx$ common from the numerator and denominator in place of $cosx$ we can also take common $sinx$ in that way we also can write the given trigonometric expression in simplest form.