If tan alpha equals the integral solution of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the inequality $4x^2 - 16x + 15 < 0$.Solving for $x$: $(2x - 3)(2x - 5) < 0$,which gives $\frac{3}{2} < x < \frac{5}{2}$.The only integer in

Vedclass · Accepted Answer

$\frac{4}{5}$

Vedclass · Answer

(D) Given the inequality $4x^2 - 16x + 15 Solving for $x$: $(2x - 3)(2x - 5) The only integer in this interval is $x = 2$. Thus,$	an \alpha = 2$.The bisector of the first quadrant is the line $y = x$,which has a slope of $1$. Thus,$\cos \beta = 1$.Using the identity $\sin(\alpha + \beta)\sin(\alpha - \beta) = \sin^2 \alpha - \sin^2 \beta$.Since $\cos \beta = 1$,we have $\sin \beta = 0$.Also,$	an \alpha = 2 \implies \sin^2 \alpha = \frac{	an^2 \alpha}{1 + 	an^2 \alpha} = \frac{2^2}{1 + 2^2} = \frac{4}{5}$.Therefore,$\sin^2 \alpha - \sin^2 \beta = \frac{4}{5} - 0 = \frac{4}{5}$.

If $\tan \alpha$ equals the integral solution of the inequality $4x^2 - 16x + 15 < 0$ and $\cos \beta$ equals the slope of the bisector of the first quadrant,then $\sin(\alpha + \beta)\sin(\alpha - \beta)$ is equal to

Similar Questions

If $\cos \theta = \frac{3}{5}$ and $\cos \phi = \frac{4}{5},$ where $\theta$ and $\phi$ are positive acute angles,then $\cos \frac{\theta - \phi}{2} = $

$\tan 15^\circ = $

$\frac{\sin 70^\circ + \cos 40^\circ}{\cos 70^\circ + \sin 40^\circ} = $

If $7 \sin a = 24 \cos a$,then the value of $14 \tan a - 75 \cos a - 7 \sec a$ is equal to:

If $\sin \beta$ is the geometric mean between $\sin \alpha$ and $\cos \alpha,$ then $\cos 2\beta$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module