^2 ( 4 - ) - ^2 ( - 4 ) = - Vidyadip

MCQ

${\cos ^2}\left( {\frac{\pi }{4} - \beta } \right) - {\sin ^2}\left( {\alpha - \frac{\pi }{4}} \right) = $

A
$\sin (\alpha + \beta )\sin (\alpha - \beta )$
B
$\cos (\alpha + \beta )\cos (\alpha - \beta )$
C
$\sin (\alpha - \beta )\cos (\alpha + \beta )$
✓
$\sin (\alpha + \beta )\cos (\alpha - \beta )$

✓

Answer

Correct option: D.

$\sin (\alpha + \beta )\cos (\alpha - \beta )$

d
(d) ${\cos ^2}\left( {\frac{\pi }{4} - \beta } \right) - {\sin ^2}\left( {\alpha - \frac{\pi }{4}} \right)$

$ = \cos \,\left( {\frac{\pi }{4} - \beta + \alpha - \frac{\pi }{4}} \right)\,\cos \,\left( {\frac{\pi }{4} - \beta - \alpha + \frac{\pi }{4}} \right)\,$

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

$= \cos (\alpha - \beta )\sin (\alpha + \beta )$.

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

Let $\tan ^{-1}(x) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, for $x \in \mathbb{R}$. Then the number of real solutions of the equation $\sqrt{1+\cos (2 x)}=\sqrt{2} \tan ^{-1}(\tan x)$ in the set $\left(-\frac{3 \pi}{2},-\frac{\pi}{2}\right) \cup\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ is equal to

Consider a sequence whose sum of first $n$ -terms is given by $S_n = 4n^2 + 6n, n \in N$, then $T_{15}$ of this sequence is -

Let $f( x)$ be a polynomial of degree four having extreme values at $ x=1 $ and $ x=2$ . If $\mathop {\lim }\limits_{x \to 0} \left[ {1 + \frac{{f\left( x \right)}}{{{x^2}}}} \right] = 3$,then $f\left( 2 \right)$ is equal to :

If the straight line $x + y = 1$ touches the parabola ${y^2} - y + x = 0$, then the co-ordinates of the point of contact are

If $X$ follows a binomial distribution with parameters $n = 6$ and $p$. If $9P\,(X = 4) = P\,(X = 2),$ then $p = $

A $2\, m$ ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate $25\, cm/ sec$., then the rate (in $cm/sec$.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is $1\, m$ above the ground is

$\int\limits_1^e {\left( {\frac{{{{\tan }^{ - 1}}x}}{x} + \frac{{\ln x}}{{\left( {1 + {x^2}} \right)}}} \right)} \,dx$ is equal to

Solve $\sin \left(\tan ^{-1} x\right),|x|<1$ is equal to

A chord $PQ$ of the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ subtends right angle at its centre. The locus of the point of intersection of tangents drawn at $P$ and $Q$ is-

let $y = y\left( x \right)$ be the solution of the differential equation $\sin x\frac{{dy}}{{dx}} + ycos\;x = 4x\;$, $x \in \left( {0,\pi } \right)$ . If $y\left( {\frac{\pi }{2}} \right) = 0$ then $y\left( {\frac{\pi }{6}} \right) = .\;.\;..\;$ .