Download App

STD 11 - 8. sequence and series — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
MCQ
If $\tan \left(\frac{\pi}{9}\right), x, \tan \left(\frac{7 \pi}{18}\right)$ are in arithmetic progression and $\tan \left(\frac{\pi}{9}\right), y, \tan \left(\frac{5 \pi}{18}\right)$ are also in arithmetic progression, then $|x-2 y|$ is equal to:
  • ✓
    $0$
  • B
    $3$
  • C
    $4$
  • D
    $1$
✓

Answer

Correct option: A.
$0$
a
$x=\frac{1}{2}\left(\tan \frac{\pi}{9}+\tan \frac{7 \pi}{18}\right)$

and $2 y=\tan \frac{\pi}{9}+\tan \frac{5 \pi}{18}$

so, $x-2 y=\frac{1}{2}\left(\tan \frac{\pi}{9}+\tan \frac{7 \pi}{18}\right)$

$-\left(\tan \frac{\pi}{9}+\tan \frac{5 \pi}{18}\right)$

$\Rightarrow|x-2 y|=\left|\frac{\cot \frac{\pi}{9}-\tan \frac{\pi}{9}}{2}-\tan \frac{5 \pi}{18}\right|$

$=\left|\cot \frac{2 \pi}{9}-\cot \frac{2 \pi}{9}\right|=0$

$\left(\operatorname{as\,\,\,\,tan} \frac{5 \pi}{18}=\cot \frac{2 \pi}{9} ; \tan \frac{7 \pi}{18}=\cot \frac{\pi}{9}\right)$

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

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

$\tan \left[ {{{\cos }^{ - 1}}\frac{4}{5} + {{\tan }^{ - 1}}\frac{2}{3}} \right] =$
→
If $\hat a,\,\hat b$ and $\hat c$ are unit vectors satisfying $\hat a\, - \,\sqrt 3 \hat b + \hat c\, = \,\vec 0,$ then the angle between the  vectors $\hat a$ and $\hat c$ is 
→
The mean of the data set comprising of $16$ observations is $16.$ If one of the observation valued $16$ is deleted and three new observations valued $3, 4$ and $5$ are added to the data, then the mean of the resultant data, is:
→
Exact set of values of $a$ for which ${x^3}(x + 1) = 2(x + a)(x + 2a)$ is having four real solutions is
→
Function $f(x) = \frac{{1 - \cos 4x}}{{8{x^2}}},$ where $x \ne 0$ and $f(x) = k$ where $x = 0$ is a continous function at $x = 0$ then the value of $k$ will be $k =........$
→
The value of $f$ at $x = 0$ so that the function $f(x) = \frac{{{2^x} - {2^{ - x}}}}{x},x \ne 0$, is continuous at $x = 0$, is
→
If $\mathop {\lim }\limits_{n \to \infty } \frac{{1 - {{(10)}^n}}}{{1 + {{(10)}^{n + 1}}}} = \frac{{ - \alpha }}{{10}}$, then give the value of $\alpha $ is
→
The number of natural number $n$ in the interval $[1005, 2010]$ for which the polynomial. $1+x+x^2+x^3+\ldots+x^{n-1}$ divides the polynomial $1+x^2+x^4+x^6+\ldots+x^{2010}$ is
→
Let ${\left( {x + 10} \right)^{50}} + {\left( {x - 10} \right)^{50}} = {a_0} + {a_1}x + {a_2}{x^2} + .... + {a_{50}}{x^{50}}$ , for $x \in R$; then $\frac{{{a_2}}}{{{a_0}}}$ is equal to
→
$\int\limits_{2 - \log 3}^{3 + \log 3} {\frac{{\log (4 + x)}}{{\log (4 + x) + \log (9 - x)}}\,\,dx = } $
→