MCQ
Write the function in the simplest form: $\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)$
  • A
    $-\frac{\pi}{4}+x$
  • B
    $-\frac{\pi}{4}-x$
  • $\frac{\pi}{4}-x$
  • D
    $\frac{\pi}{4}+x$

Answer

Correct option: C.
$\frac{\pi}{4}-x$
c
$\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)$

$=\tan ^{-1}\left(\frac{1-\left(\frac{\sin x}{\cos x}\right)}{1+\left(\frac{\sin x}{\cos x}\right)}\right)$

$=\tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)$

$=\tan ^{-1}(1)-\tan ^{-1}(\tan x)$ $\left[\because \frac{-y}{x y}=\tan ^{-1} x-\tan ^{-1} y\right]$

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

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

$\int\limits_{\frac{\pi }{6}}^{\frac{{5\pi }}{6}} {\left( {\frac{1}{2}{{(3\sin \theta )}^2} - \frac{1}{2}{{(1 + \sin \theta )}^2}} \right)\,d\theta } $
If f : [1, $\infty$) → [2, $\infty$) is given by $\text{f(x)}=\text{x}+\frac{1}{\text{x}},$ then f-1 equals to:
  1. $\frac{\text{x}+\sqrt{\text{x}^2-4}}{2}$
  2. $\frac{\text{x}}{1+\text{x}^2}$
  3. $\frac{\text{x}-\sqrt{\text{x}^2-4}}{2}$
  4. $1+\sqrt{\text{x}^2-4}$
$f(x) = \left\{ {\begin{array}{*{20}{c}}
  {\left[ {\cos \pi x} \right];\,\,\,\,x \leqslant 1} \\ 
  {2\left\{ x \right\} - 1\,\,\,\,x > 1} 
\end{array}} \right.$  comment on the derivability at $x =1,$ where $[ \ ]$ denotes greatest integer function $and\, \{ \}$ denotes fractional part function
The area of the region bounded by the curve $y=\cos x, x=0$ and $x=\frac{3 \pi}{2}$ is __________ .
If $AB = A$ and $BA = B$, then
If $\text{f(x)}=\begin{cases}\text{mx}+1,&\text{x}\leq\frac{\pi}{2}\\\sin\text{x}+\text{n},&\text{n}>\frac{\pi}{2}\end{cases}$ is continuous at $\text{x}=\frac{\pi}{2},$ then:
  1. $\text{m}=1,\text{ n}=0$
  2. $\text{m}=\frac{\text{n}\pi}{2}+1$
  3. $\text{n}=\frac{\text{m}\pi}{2}$
  4. $\text{m}=\text{n}=\frac{\pi}{2}$
The principal solution of $\cos ^{-1}\left(\cos \left(\frac{7 \pi}{6}\right)\right)$ is
The value of $\int_0^{\pi /4} {\frac{{1 + \tan x}}{{1 - \tan x}}\,dx} $ is
If the two lines $l_{1}: \frac{ x -2}{3}=\frac{ y +1}{-2}, z =2$ and $l_{2}: \frac{x-1}{1}=\frac{2 y+3}{\alpha}=\frac{z+5}{2}$ perpendicular, then an angle between the lines $l_{2}$ and $l_{3}: \frac{1- x }{3}=\frac{2 y -1}{-4}=\frac{ z }{4}$ is
$\int\limits^\frac{\pi}{2}_0\frac{1}{2+\cos\text{x}}\text{ dx}$ equals:
  1. $\frac{1}{3}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
  2. $\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
  3. ${\sqrt{3 }}\tan^{-1}\big({\sqrt{3}}\big)$
  4. $2{\sqrt{3 }}\tan^{-1}{\sqrt{3}}$