MCQ
${\tan ^{ - 1}}\left( {\frac{1}{4}} \right) + {\tan ^{ - 1}}\left( {\frac{2}{9}} \right) = $
  • $\frac{1}{2}{\cos ^{ - 1}}\left( {\frac{3}{5}} \right)$
  • B
    $\frac{1}{2}{\sin ^{ - 1}}\left( {\frac{3}{5}} \right)$
  • C
    $\frac{1}{2}{\tan ^{ - 1}}\left( {\frac{3}{5}} \right)$
  • D
    None of these

Answer

Correct option: A.
$\frac{1}{2}{\cos ^{ - 1}}\left( {\frac{3}{5}} \right)$
a
(a) ${\tan ^{ - 1}}\frac{1}{4} + {\tan ^{ - 1}}\frac{2}{9} = {\tan ^{ - 1}}\left( {\frac{{(1/4) + (2/9)}}{{1 - (1/4) \times (2/9)}}} \right)$

$={\tan ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{1}{2}.2{\tan ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{1}{2}{\tan ^{ - 1}}\frac{{2(1/2)}}{{1 - (1/4)}}$

$ = \frac{1}{2}{\tan ^{ - 1}}\frac{4}{3} = \frac{1}{2}{\cos ^{ - 1}}\frac{3}{5}$.

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

The number of arbitrary constants in the particular solution of a differential equation of third order are:
The value of $\left|\begin{array}{lll}(a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1\end{array}\right|$ is 
Let $L_1: \overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}), \lambda \in \mathrm{R}$ $L_2: \overrightarrow{\mathrm{r}}=(\hat{\mathrm{j}}-\hat{\mathrm{k}})+\mu(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\mathrm{p} \hat{\mathrm{k}}), \mu \in \mathrm{R}$ and  $L_3: \overrightarrow{\mathrm{r}}=\delta(\ell \hat{\mathrm{i}}+\mathrm{m} \hat{\mathrm{j}}+\mathrm{n} \hat{\mathrm{k}}) \delta \in \mathrm{R}$

Be three lines such that $\mathrm{L}_1$ is perpendicular to $\mathrm{L}_2$ and $L_3$ is perpendicular to both $L_1$ and $L_2$. Then the point which lies on $\mathrm{L}_3$ is

The integrating factor of the differential equation $\left(x+3 y^2\right) \frac{d y}{d x}=y$ is
If $f : R \rightarrow R$ defind by $\text{f(x)}=\frac{2\text{x}-7}{4}$ is an invertible function, then find $f^{-1}.$
Choose the correct answer from the given four options:
The sides of an equilateral triangle are increasing at the rate of 2cm/ sec. The rate at which the area increases, when side is 10cm is:
If $f(x) = \left\{ \begin{array}{l}x + \lambda ,\;x\, < 3\\\,\,\,\,\,\,\,\,\,4,\,\,x = 3\\3x - 5,\,\,x > 3\end{array} \right.$ is continuous at $x = 3$, then $\lambda = $
$\mathop {Lim}\limits_{\lambda  \to 0} \,\,{\left( {\int\limits_0^1 {{{(1 + x)}^\lambda }dx} } \right)^{\frac{1}{\lambda }}}$ is equal to
If $\text{y}=\text{a}\cos(\log_\text{e}\text{x})+\text{b}\sin(\log_\text{e}\text{x})$ then $\text{x}^2\text{y}^2+\text{xy}_1=$
If function $f$ is defined such that
$
f(x)=\left\{\begin{array}{cc}
\frac{k \cos x}{\pi-2 x}, & \text { if } x \neq \frac{\pi}{2} \\
3, & \text { if } x=\frac{\pi}{2}
\end{array}\right.
$is continuous at $x=\frac{\pi}{2}$, then value of $k$ :