Let alpha = 3 sin^{-1}(frac{6}{11}) and beta | vedclass

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

Question

(A) Given $\alpha = 3 \sin^{-1}(\frac{6}{11})$. Since $\frac{6}{11} > \frac{1}{2}$, we have $\sin^{-1}(\frac{6}{11}) > \sin^{-1}(\frac{1}{2}) = 30^\ci

Vedclass · Accepted Answer

Both Statement $I$ and Statement $II$ are true

Vedclass · Answer

(A) Given $\alpha = 3 \sin^{-1}(\frac{6}{11})$. Since $\frac{6}{11} > \frac{1}{2}$, we have $\sin^{-1}(\frac{6}{11}) > \sin^{-1}(\frac{1}{2}) = 30^\circ$. Thus, $\alpha > 3 	imes 30^\circ = 90^\circ$. Since $90^\circ Given $\beta = 3 \cos^{-1}(\frac{4}{9})$. Since $\frac{4}{9}  \cos^{-1}(\frac{1}{2}) = 60^\circ$. Thus, $\beta > 3 	imes 60^\circ = 180^\circ$. Since $\alpha > 90^\circ$ and $\beta > 180^\circ$, $\alpha + \beta > 270^\circ$. In the fourth quadrant, the cosine function is positive. Therefore, $\cos(\alpha + \beta) > 0$. So, Statement $I$ is true.

Similar Questions

The solution set of the inequality $(\tan^{-1} x)(\cot^{-1} x) - (\tan^{-1} x)(1 + \frac{\pi}{2}) - 2\cot^{-1} x + 2(1 + \frac{\pi}{2}) > \lim_{x \to \infty} [\sec^{-1} x - \frac{\pi}{2}]$ is (where $[.]$ denotes the greatest integer function):

$\sin \left[ \cos^{-1} \left( \frac{3}{5} \right) + \tan^{-1} 2 \right] = $

$\tan \left[\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]+\tan \left[\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right]$ is equal to

Evaluate the sum: $\sum_{k=1}^{2026} \sin^{-1}(\cos \frac{k\pi}{4})$.

If we consider only the principal values of the inverse trigonometric functions,then the value of $\tan \left( \cos^{-1} \frac{1}{5\sqrt{2}} - \sin^{-1} \frac{4}{\sqrt{17}} \right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module