Given that ( - 2 ) = 2 ( + B 2 ), then 2 2 is equal to - Vidyadip

MCQ

Given that $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = 2\cos \left( {\frac{{\alpha + B}}{2}} \right)$, then $\tan \frac{\alpha }{2}\tan \frac{\beta }{2} $ is equal to

A
$\frac{1}{2}$
✓
$1\over3$
C
$\frac{1}{4}$
D
$\frac{1}{8}$

✓

Answer

Correct option: B.

$1\over3$

b
(b) $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = 2\cos \left( {\frac{{\alpha + \beta }}{2}} \right)$

==> $\cos \frac{\alpha }{2}\cos \frac{\beta }{2} + \sin \frac{\alpha }{2}\sin \frac{\beta }{2} = 2\cos \frac{\alpha }{2}\cos \frac{\beta }{2} - 2\sin \frac{\alpha }{2}\sin \frac{\beta }{2}$

$ \Rightarrow 3\sin \frac{\alpha }{2}\sin \frac{\beta }{2} = \cos \frac{\alpha }{2}\cos \frac{\beta }{2}$

==> $\tan \frac{\alpha }{2}\tan \frac{\beta }{2} = \frac{1}{3}$.

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

If complex number $z = x + iy$ is taken such that the amplitude of fraction $\frac{{z - 1}}{{z + 1}}$ is always $\frac{\pi }{4}$, then

A spherical iron ball of radius $10\,cm$ is coated with a layer of ice of uniform thickness that melts at a rate of $50\,cm^3/min.$ When the thickness of the ice is $5\,cm,$ then the rate at which the thickness (in $cm/min$ ) of ice decreases is

$\sin 4\theta $ can be written as

Let $\mathrm{S}$ be the set of positive integral values of $a$ for which $\frac{\mathrm{ax}^2+2(\mathrm{a}+1) \mathrm{x}+9 \mathrm{a}+4}{\mathrm{x}^2-8 \mathrm{x}+32}<0, \forall \mathrm{x} \in \mathbb{R}$. Then, the number of elements in $\mathrm{S}$ is :

The equation of the circle which passes through the point of intersection of circles ${x^2} + {y^2} - 8x - 2y + 7 = 0$ and ${x^2} + {y^2} - 4x + 10y + 8 = 0$ and having its centre on $y$ - axis, will be

Let $A=\left[\begin{array}{cc}\alpha & -1 \\ 6 & \beta\end{array}\right], \alpha>0$, such that $\operatorname{det}(A)=0$ and $\alpha+\beta=1$. If I denotes $2 \times 2$ identity matrix, then the matrix $(1+ A )^8$ is:

The slope of the tangent to a curve $C : y = y ( x )$ at any point $[ x , y )$ on it is $\frac{2 e ^{2 x }-6 e ^{- x }+9}{2+9 e ^{-2 x }}$. If $C$ passes through the points $\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)$ and $\left(\alpha, \frac{1}{2} e ^{2 \alpha}\right)$ then $e ^{\alpha}$ is equal to.

For all values of $\theta $, the locus of the point of intersection of the lines $x\cos \theta + y\sin \theta = a$ and $x\sin \theta - y\cos \theta = b$ is

The sum of first four terms of a geometric progression $(G.P.)$ is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18} .$ If the product of first three terms of the $G.P.$ is $1,$ and the third term is $\alpha$, then $2 \alpha$ is ....... .

Consider the lines $\mathrm{L}_{1}: \mathrm{x}-1=\mathrm{y}-2=\mathrm{z}$ and $\mathrm{L}_{2}: \mathrm{x}-2=\mathrm{y}=\mathrm{z}-1$. Let the feet of the perpendiculars from the point $\mathrm{P}(5,1,-3)$ on the lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ be Q and R respectively. If the area of the triangle $P Q R$ is $A$, then $4 A^{2}$ is equal to :