Question
Choose the correct answer from the given four options.
The value of $\sin\big[2\tan^{-1}(0.75)\big]$ is equal to:
  1. 0.75
  2. 1.5
  3. 0.96
  4. sin1.5

Answer

  1. 0.96

Solution:

We have, $\sin\big[2\tan^{-1}(0.75)\big]$

$=\sin\Big(2\tan^{-1}\frac{3}{4}\Big)$

$=\sin\Bigg(\sin^{-1}\frac{2.\frac{3}{4}}{1+\frac{9}{16}}\Bigg)$

$\Big(\because\ 2\tan^{-1}\text{x}=\sin^{-1}\frac{2\text{x}}{1+\text{x}^2}\Big)$

$=\sin\Bigg(\sin^{-1}\frac{\frac{3}{2}}{\frac{25}{16}}\Bigg)$

$=\sin\Big(\sin^{-1}\frac{24}{25}\Big)=\frac{24}{25}=0.96$

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

$\int_{}^{} {\frac{{\sin x\;dx}}{{{a^2} + {b^2}{{\cos }^2}x}}} = $
If the events $A$ and $B$ are mutually exclusive, then $P\left( {\frac{A}{B}} \right) = $
Vector coplanar with vectors $i + j $ and $ j + k$  and parallel to the vector $2i -2j -4k$ , is
Consider a LPP given by
Minimum Z = 6x + 10y
Subjected to x ≥ 6, y ≥ 2, 2x + y ≥ 10, x ≥ 0, y ≥ 0
Redundant constraints in this LPP are
  1. x ≥ 0, y ≥ 0
  2. x ≥ 6
  3. 2x + y ≥ 10
  4. none of these
If $f(x) = \left\{ \begin{array}{l}\frac{{{x^4} - 16}}{{x - 2}},\,\,{\rm{when}}\,\,x \ne 2\\\,\,\,\,\,\,\,\,\,\,\,\,\,16,\,{\rm{when}}\,\,x = 2\end{array} \right.$, then
If matrix A is of order p × q and matrix B is of order r × s then  A − B will exist if:
  1. p = q
  2. p = r, q = s
  3. p = q, r = s
  4. p = s, q = r
If $A=\left[\tan \left(\frac{\theta}{2}\right)^{-\tan \left(\frac{\theta}{2}\right)}{0}\right]$ and $\left( I _{2}+ A \right)\left( I _{2}- A \right)^{-1}=\left[\begin{array}{ll} a & - b \\ b & a \end{array}\right],$ then $13\left( a ^{2}+ b ^{2}\right)$ is equal to ...........
$A = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&1\\0&{ - 2}&4\end{array}} \right];\,\,I = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$${A^{ - 1}} = \frac{1}{6}[{A^2} + cA + dI]$ where $c,d \in R$, then pair of values $(c,d)$
Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}=2 x(x+y)^3-x(x+y)-1, y(0)=1$. Then, $\left(\frac{1}{\sqrt{2}}+\mathrm{y}\left(\frac{1}{\sqrt{2}}\right)\right)^2$ equals : . . . .  . . .
$f(x,\;y) = \frac{1}{{x + y}}$ is a homogeneous function of degree