Download App

STD 12 - 2. inverse trigonometric function — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 2. inverse trigonometric function1 Mark
MCQ
$4{\tan ^{ - 1}}\frac{1}{5} - {\tan ^{ - 1}}\frac{1}{{239}}$ is equal to
  • A
    $\pi $
  • B
    $\frac{\pi }{2}$
  • C
    $\frac{\pi }{3}$
  • ✓
    $\frac{\pi }{4}$
✓

Answer

Correct option: D.
$\frac{\pi }{4}$
d
(d) Since $2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}$

$\therefore$ $4{\tan ^{ - 1}}\frac{1}{5} = 2\,\left[ {2{{\tan }^{ - 1}}\frac{1}{5}} \right] = 2{\tan ^{ - 1}}\frac{{\frac{2}{5}}}{{1 - \frac{1}{{25}}}}$

$ = 2{\tan ^{ - 1}}\frac{{10}}{{24}} = {\tan ^{ - 1}}\frac{{\frac{{20}}{{24}}}}{{1 - \frac{{100}}{{576}}}} = {\tan ^{ - 1}}\frac{{120}}{{119}}$

So, $4{\tan ^{ - 1}}\frac{1}{5} - {\tan ^{ - 1}}\frac{1}{{239}} = {\tan ^{ - 1}}\frac{{120}}{{119}} - {\tan ^{ - 1}}\frac{1}{{239}}$

$ = {\tan ^{ - 1}}\frac{{\frac{{120}}{{119}} - \frac{1}{{239}}}}{{1 + \frac{{120}}{{119}}.\frac{1}{{239}}}} = {\tan ^{ - 1}}\frac{{(120 \times 239) - 119}}{{(119 \times 239) + 120}}$

==> ${\tan ^{ - 1}}1 = \frac{\pi }{4}$.

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

If $f(x + y) = f(x).f(y)$ for all $x$ and $y$ and $f(5) = 2$, $f'(0) = 3$, then $f'(5)$ will be
→
Area bounded by the curve ${x^2} = 4y$ and the straight line $x = 4y - 2$ is
→
If $\vec{a}$ is a nonzero vector of magnitude ' $a$ ' and $\lambda$ a nonzero scalar, then $\lambda \vec{a}$ is unit vector if :
→
Let ' $a$ ' be a real number such that the function $f(x)=a x^{2}+6 x-15, x \in R$ is increasing in $\left(-\infty, \frac{3}{4}\right)$ and decreasing in $\left(\frac{3}{4}, \infty\right) .$ Then the function $g(x)=a x^{2}-6 x+15, x \in R$ has a:
→
The order of the differential equation of all circles of radius $r$, having centre on $y$-axis and passing through the origin is
→
$\int_{}^{} {\sec x\log (\sec x + \tan x)\;dx = } $
→
$\int {\frac{{3\cos x + 2\sin x}}{{4\sin x + 5\cos x}}dx = }\, A \{23x + 2f(4 \sin\, x + 5\, \cos \, x)\} + c,$ then $A$ and $f(x)$ are
→
Line passing through $(3, 4, 5)$ and $(4, 5, 6)$ has direction ratios:
→
$\int_{1/4}^{1/2} {\frac{{dx}}{{\sqrt {x - {x^2}} }} = } $
→
A tangent having slope of $-\frac{4}{3}$ to the ellipse $\frac{\text{x}^2}{18}+\frac{\text{y}^2}{32}=1$ ntersects the major and minor axes at points $A$ and $B$ respectively. If $C$ is the center of the ellipse, then area of the triangle $\text{ABC}$ is:
→