If frac{x}{cos alpha} = frac{y}{cos left(frac | vedclass

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

Question

(B) Let $\frac{x}{\cos \alpha} = \frac{y}{\cos \left(\frac{2 \pi}{3} - \alpha\right)} = \frac{z}{\cos \left(\frac{2 \pi}{3} + \alpha\right)} = k$. The

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Let $\frac{x}{\cos \alpha} = \frac{y}{\cos \left(\frac{2 \pi}{3} - \alpha\right)} = \frac{z}{\cos \left(\frac{2 \pi}{3} + \alpha\right)} = k$. Then,$x = k \cos \alpha$,$y = k \cos \left(\frac{2 \pi}{3} - \alpha\right)$,and $z = k \cos \left(\frac{2 \pi}{3} + \alpha\right)$. Now,$x + y + z = k \left[ \cos \alpha + \cos \left(\frac{2 \pi}{3} - \alpha\right) + \cos \left(\frac{2 \pi}{3} + \alpha\right) \right]$. Using the formula $\cos(A - B) + \cos(A + B) = 2 \cos A \cos B$,we get: $x + y + z = k \left[ \cos \alpha + 2 \cos \left(\frac{2 \pi}{3}\right) \cos \alpha \right]$. Since $\cos \left(\frac{2 \pi}{3}\right) = -\frac{1}{2}$,we have: $x + y + z = k \left[ \cos \alpha + 2 \left(-\frac{1}{2}\right) \cos \alpha \right] = k [\cos \alpha - \cos \alpha] = 0$.

If $\frac{x}{\cos \alpha} = \frac{y}{\cos \left(\frac{2 \pi}{3} - \alpha\right)} = \frac{z}{\cos \left(\frac{2 \pi}{3} + \alpha\right)}$,then the value of $(x + y + z)$ is equal to

Similar Questions

Prove that: $2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13} + \cos \frac{3 \pi}{13} + \cos \frac{5 \pi}{13} = 0$

For $\theta \in \left(0, \frac{\pi}{2}\right)$,if $\tan 3\theta \cdot \tan 2\theta \cdot \tan \theta + \tan 2\theta + \tan \theta = 1$,then $\theta =$

In an isosceles right-angled triangle,a straight line is drawn from the midpoint of one of the equal sides to the opposite vertex. Then a pair of possible values of the cotangents of the two angles so formed at that vertex are

$\tan 5x \tan 3x \tan 2x = $

Find the value of: $\sin 75^{\circ}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module