MCQ
If ${\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}A,$ then $A$
  • A
    $x - y$
  • B
    $x + y$
  • $\frac{{x - y}}{{1 + xy}}$
  • D
    $\frac{{x + y}}{{1 - xy}}$

Answer

Correct option: C.
$\frac{{x - y}}{{1 + xy}}$
c
(c) Given that ${\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}A$

$ \Rightarrow \,\,{\tan ^{ - 1}}\,\left( {\frac{{x - y}}{{1 + xy}}} \right) = {\tan ^{ - 1}}A$.

Hence $A = \frac{{x - y}}{{1 + xy}}$.

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

Consider the function $\mathrm{f}:\left[\frac{1}{2}, 1\right] \rightarrow \mathrm{R}$ defined by $f(x)=4 \sqrt{2} x^3-3 \sqrt{2} x-1$. Consider the statements

$(I)$ The curve $y=f(x)$ intersects the $x$-axis exactly at one point

$(II)$ The curve $y=f(x)$ intersects the $x$-axis at $\mathrm{x}=\cos \frac{\pi}{12}$

Then

Let the function f : R - {-b} → R - {1} be defined by $\text{f(x)}=\frac{\text{x}+\text{a}}{\text{x}+\text{b}},\ \text{a}\neq\text{b}.$ Then,
Let $f(x)$ = $\int\limits_0^x {({t^2} + 2t + 2)dt} $ where $x$ is set of real numbers satisfying the inequation ${\log _{\sqrt 2 }}(1 + \sqrt {6x - {x^2} - 8} ) \ge 0$ If range of $f(x)$ is $[a, b]$ then $(a + b)$ is
The area of a parallelogram whose two adjacent sides are represented by the vector $3i - k$ and $i + 2j$ is
Given $\int\limits_0^{\frac{\pi }{2}} {\,\,\frac{{dx}}{{1 + \sin x + \cos x}}}  = ln\, 2$, then the value of the def. integral. $\int\limits_0^{\frac{\pi }{2}} {\,\,\frac{{\sin \,x}}{{1 + \sin x + \cos x}}} \,dx$ is equal to
If $\text{A}=\begin{bmatrix}\alpha &\text{amp; 2} \\2 &\text{amp; }\alpha \end{bmatrix}$and $ |\text{A}^3|=125,$ then $\alpha$ is equal to:
The direction cosines of a line equally inclined to three mutually perpendicular lines having direction cosines as ${l_1},{m_1},{n_1};{l_2},{m_2},{n_2}$ and ${l_3},{m_3},{n_3}$ are
The equation of the curve passing through the origin and satisfying the equation $(1 + {x^2})\frac{{dy}}{{dx}} + 2xy = 4{x^2}$ is
Find the area of the region bounded by the curves $y = x^3,$ the line $x = 2, x = 5$ and the $x -$ axis?
$\int\frac{\text{dx}}{1+\text{\cos x}}=$