Download App

Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions1 Mark
MCQ
The value of $\Big(\cot\frac{\text{x}}{2}-\tan\frac{\text{x}}{2}\Big)^2(1-2\tan\text{x}\cot2\text{x})$ is:
  • A
    $1$
  • B
    $2$
  • C
    $3$
  • ✓
    $4$
✓

Answer

Correct option: D.
$4$
We have,
$\big(\cot\frac{\text{x}}{2}-\tan\frac{\text{x}}{2}\big)(1-2\tan\text{x}\cot2\text{x})$
$\big(\cot^2\frac{\text{x}}{2}-2\cot\frac{\text{x}}{2}+\tan^2\frac{\text{x}}{2}\big)\Big\{1-2\tan\text{x}\Big(\frac{\cot^2\text{x}-1}{2\cot\text{x}}\Big)\Big\}$
$\big(\cot^2\frac{\text{x}}{2}-2+\tan^2\frac{\text{x}}{2}\big)\Big\{1-\tan\text{x}\Big(\frac{\cot^2\text{x}-1}{\cot\text{x}}\Big)\Big\}$
$\big(\cot^2\frac{\text{x}}{2}+\tan^2\frac{\text{x}}{2}-2\big)\Big(1-\frac{\cot^2\text{x}-\tan\text{x}}{\cot\text{x}}\Big)$
$\big(\cot^2\frac{\text{x}}{2}+\tan^2\frac{\text{x}}{2}-2\big)\big(\tan^2\text{x}\big)$
$\big(\cot^2\frac{\text{x}}{2}+\tan\frac{\text{x}}{2}-2\big)\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1-\tan^2\frac{\text{x}}{2}}\Bigg)^2$
$=\frac{1}{\big(1-\tan^2\frac{\text{x}}{2}\big)^2}\big(4+4\tan^4\frac{\text{x}}{2}-8\tan^2\frac{\text{x}}{2}\big)$
$=\frac{1}{\big(1-\tan^2\frac{\text{x}}{2}\big)^2}\big(4-8\tan^2\frac{\text{x}}{2}+4\tan^4\frac{\text{x}}{2}\big)$
$=\frac{1}{\big(1-\tan^2\frac{\text{x}}{2}\big)^2}\Big\{\big(\tan^2\frac{\text{x}}{2}\big)^2-2\big(\tan^2\frac{\text{x}}{2}+1\big)\Big\}$
$=\frac{4\big(\tan^2\frac{​​\text{x}}{2}-1\big)^2}{\big(1-\tan^2\frac{\text{x}}{2}\big)^2}$
$=4$

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 Trigonometric Functions→Trigonometric Functions — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

Statement $(a+i b)<(c+i d)$ is true for which of the followings :
→
If $\cot (\alpha + \beta ) = 0,$ then $\sin (\alpha + 2\beta ) = $
→
Let A and B be two sets in the same universal set. Then, A - B =
→
A man starts walking from the point $\mathrm{P}(-3,4)$, touches the $\mathrm{x}$-axis at $\mathrm{R}$, and then turns to reach at the point $\mathrm{Q}(0,2) .$ The man is walking at a constant speed. If the man reaches the point $Q$ in the minimum time, then $50\,\left((\mathrm{PR})^{2}+(\mathrm{RQ})^{2}\right)$ is equal to ..... .
→
Two dice are thrown independently. Let $A$ be the event that the number appeared on the $1^{\text {st }}$ die is less than the number appeared on the $2^{\text {nd }}$ die, $B$ be the event that the number appeared on the $1^{\text {st }}$ die is even and that on the second die is odd, and $C$ be the event that the number appeared on the $1^{\text {st }}$ die is odd and that on the $2^{\text {nd }}$ is even. Then
→
The integral values of $a$ for which the quadratic equation $(x - a) (x - 10) + 1 = 0$ has integral roots are
→
If $f : [-2, 2] \rightarrow R$ is defined by $\text{f(x)}=\begin{cases}-1,&\text{for}-2\leq\text{x}\leq0\\\text{x}-1,&\text{for }0\leq\text{x}\leq2\end{cases}$ then $\{\text{x}\in[-2,2]:\text{x}\leq0\text{ and }\text{f}(\text{|x|})=\text{x}\}=$
→
The partial fractions of ${{3{x^3} - 8{x^2} + 10} \over {{{(x - 1)}^4}}}$ is
→
The points $\left( {\frac{a}{{\sqrt 3 }},a} \right),\;\left( {\frac{{2a}}{{\sqrt 3 }},\,2a} \right),\;\left( {\frac{a}{{\sqrt 3 }},\,3a} \right)$ are the vertices of
→
If in an $A.P.,$ first term is $20,$ common difference is $2$ and nth term is $42$, then find $n.$
→