If ^ - 1 ( 1 x ) = , then = - Vidyadip

MCQ

If ${\cos ^{ - 1}}\left( {\frac{1}{x}} \right) = \theta $, then $\tan \theta =$

A
$\frac{1}{{\sqrt {{x^2} - 1} }}$
B
$\sqrt {{x^2} + 1} $
C
$\sqrt {1 - {x^2}} $
✓
$\sqrt {{x^2} - 1} $

✓

Answer

Correct option: D.

$\sqrt {{x^2} - 1} $

d
(d) Given that ${\cos ^{ - 1}}\left( {\frac{1}{x}} \right) = \theta \,\, \Rightarrow \,\,\cos \theta = \frac{1}{x}$

Now, $\tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{{\sqrt {1 - {{(1/x)}^2}} }}{{1/x}} = \sqrt {{x^2} - 1} $.

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 STD 12 - 2. inverse trigonometric function→STD 12 - 2. inverse trigonometric function — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

The value of $\cos ({\tan ^{ - 1}}(\tan 2))$ is

$\cot(\frac{\pi}{4}-2\cot^{-1}3)=$

The value of $\cot \left( {\sum\limits_{r = 1}^\infty {{{\tan }^{ - 1}}\left( {\frac{4}{{4{r^2} + 3}}} \right)} } \right)$ is equal to

Let $\vec u\;$be a vector coplanar with the vector $\vec a = 2\hat i + 3\hat j - \hat k$ and $\vec b = \hat j + \hat k$ . If $\vec u$ is perpendicular to $\vec a$ and $\vec u \cdot \vec b = 24$ ,then ${\left| {\vec u} \right|^2} = $ . . . .

Let the value of $p = (x + 4y)\,a + (2x + y + 1)\,b$ and $q = (y - 2x + 2)\,a + (2x - 3y - 1)\,b,$ where $a $ and $b $ are non-collinear vectors. If $3p = 2q,$ then the value of $x $ and $y$ will be

Let $y\,(x)$ be a solution of $\frac{{(2 + \sin \,x)\,dy}}{{(1 + y)dx}} = \cos \,\,x.$ If $y(0) = 2,$ then $y\left( {\frac{\pi }{2}} \right)$ equals

Let $N$ denote the set of all natural numbers. Define two binary relations on $N$ as $R_1 = \{(x,y) \in N \times N : 2x + y= 10\}$ and $R_2 = \{(x,y) \in N\times N : x+ 2y= 10\} $. Then

If there is an error of 2% in measuring the length of a simple pendulum, then percentage error in its period is:

Choose the correct answer from the given four options.
If $|\text{x}|\leq1,$ then $2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ is equal to:

The solution of ${y^5}x + y - x\frac{{dy}}{{dx}} = 0$ is