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 $\cos A = m \cos B,$ then

If $2\tan A = 3\tan B,$ then $\frac{\sin 2B}{5 - \cos 2B}$ is equal to

If $\tan \theta \cdot \tan 2 \theta = 1,$ then the value of $\sin ^{2} 2 \theta + \tan ^{2} 2 \theta$ is equal to

If $\cos (\theta - \alpha ), \cos \theta$ and $\cos (\theta + \alpha )$ are in $H.P.$,then $\cos \theta \sec \frac{\alpha }{2}$ is equal to

If $\log_a b + \log_b c + \log_c a$ vanishes,where $a, b,$ and $c$ are positive real numbers different from unity,then the value of $(\log_a b)^3 + (\log_b c)^3 + (\log_c a)^3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module