Download App

2. inverse trigonometric function — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths2. inverse trigonometric function1 Mark
MCQ
${\tan ^{ - 1}}\frac{{1 - {x^2}}}{{2x}} + {\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}} = $
  • A
    $\frac{\pi }{4}$
  • ✓
    $\frac{\pi }{2}$
  • C
    $\pi $
✓

Answer

Correct option: B.
$\frac{\pi }{2}$
b
(b) Putting $x = \tan \theta $

${\tan ^{ - 1}}\frac{{1 - {x^2}}}{{2x}} + {\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}}$

$ = {\tan ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{2\tan \theta }}} \right) + {\cos ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$ = {\tan ^{ - 1}}(\cot 2\theta ) + {\cos ^{ - 1}}(\cos 2\theta )$

$ = \frac{\pi }{2} - 2\theta + 2\theta = \frac{\pi }{2}$.

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

Explore more

More "M.C.Q (1 Marks)" from 2. inverse trigonometric function→2. inverse trigonometric function — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Area under the curve $y = \sin 2x + \cos 2x$ between $x = 0$ and $x = \frac{\pi }{4},$ is ......... $sq. \,unit$
→
Find the direction cosines of the line $\frac{x-2}{2}=\frac{2 y-5}{-3}=\frac{z+1}{0}$.
→
Let $f:\left( { - 1,1} \right) \to R$ be differentiable function with $f\left( 0 \right) = - 1$ and $f'\left( 0 \right) = 1$ . Let $g\left( x \right) = {\left[ {f\left( {2f\left( x \right) + 2} \right)} \right]^2}$ .Then $g'\left( 0 \right) = $
→
Let  $h(x)$ be differentiable for all $x$ and let $f(x) = (kx  + e^x)$  $h(x)$ . Where $k$  is some constant If $h(0) = 5 , h'(0) = -2$ & $f'(0) = 18$ then the value of $k$ is equal to
→
$\int_0^{2 \pi} \sin ^3 x \cos ^2 x d x=$ ___________ .
→
The value of $\big(\vec{\text{a}}\times\vec{\text{b}}\big)^2$ is:
  1. $|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
  2. $|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2-\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$
  3. $|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\big(\vec{\text{a}}.\vec{\text{b}}\big)$
  4. $|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-\vec{\text{a}}.\vec{\text{b}}$
→
Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than $99\%$ is
→
There are three bags $X$, $Y$ and $Z$. Bag $X$ contains $5$ one-rupee coins and $4$ five-rupee coins; Bag $\mathrm{Y}$ contains $4$ one-rupee coins and $5$ five-rupee coins and Bag $\mathrm{Z}$ contains $3$ one-rupee coins and $6$ five-rupee coins. A bag is selected at random and a coin drawn from it at random is found to be a one-rupee coin. Then the probability, that it came from bag $\mathrm{Y}$, is :
→
A quadratic polynomial $P(x)$ satisfies the conditions, $P(0) = P(1) = 0\, \&\,\int\limits_0^1 {}  P(x) dx = 1$. The leading coefficient of the quadratic polynomial is :
→
The general solution of the differential equation $\frac{d y}{d x}+\frac{y}{x}=1$ is
→