If cos ^{-1} alpha+cos ^{-1} beta+cos ^{-1} g | vedclass

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

Question

(C) Given that $\cos ^{-1} \alpha+\cos ^{-1} \beta+\cos ^{-1} \gamma=3 \pi$. We know that the range of $\cos ^{-1} x$ is $[0, \pi]$. Since the sum of

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) Given that $\cos ^{-1} \alpha+\cos ^{-1} \beta+\cos ^{-1} \gamma=3 \pi$. We know that the range of $\cos ^{-1} x$ is $[0, \pi]$. Since the sum of three values, each at most $\pi$, is $3 \pi$, each term must be equal to $\pi$. Therefore, $\cos ^{-1} \alpha = \pi$, $\cos ^{-1} \beta = \pi$, and $\cos ^{-1} \gamma = \pi$. This implies $\alpha = \cos(\pi) = -1$, $\beta = \cos(\pi) = -1$, and $\gamma = \cos(\pi) = -1$. Now, we calculate the expression $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$. Substituting $\alpha = -1$, $\beta = -1$, and $\gamma = -1$: $(-1)(-1-1) + (-1)(-1-1) + (-1)(-1-1) = (-1)(-2) + (-1)(-2) + (-1)(-2) = 2 + 2 + 2 = 6$.

If $\cos ^{-1} \alpha+\cos ^{-1} \beta+\cos ^{-1} \gamma=3 \pi$, then $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$ is equal to

Similar Questions

The value of $2 \cot ^{-1} \frac{1}{2} - \cot ^{-1} \frac{4}{3}$ is

If in a triangle $ABC$,$A = \tan^{-1} 2$ and $B = \tan^{-1} 3$,then angle $C$ is equal to

If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$, then

Considering the principal values of the inverse trigonometric functions,$\sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^2}\right)$,for $-\frac{1}{2} < x < \frac{1}{\sqrt{2}}$,is equal to:

In a $\triangle ABC$,if $\angle A = 90^{\circ}$,then $\cos^{-1}\left(\frac{R}{r_2+r_3}\right)$ is equal to (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module