If frac{1}{6} sin theta, cos theta, tan theta | vedclass

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

Question

(A) Given that $\frac{1}{6} \sin 	heta, \cos 	heta, 	an 	heta$ are in $G.P.$Therefore,$(\cos 	heta)^2 = (\frac{1}{6} \sin 	heta) 	imes (	an

Vedclass · Accepted Answer

$2 n \pi \pm \frac{\pi}{3}, n \in Z$

Vedclass · Answer

(A) Given that $\frac{1}{6} \sin 	heta, \cos 	heta, 	an 	heta$ are in $G.P.$Therefore,$(\cos 	heta)^2 = (\frac{1}{6} \sin 	heta) 	imes (	an 	heta)$$\cos^2 	heta = \frac{1}{6} \sin 	heta 	imes \frac{\sin 	heta}{\cos 	heta}$$\cos^3 	heta = \frac{1}{6} \sin^2 	heta$$\cos^3 	heta = \frac{1}{6} (1 - \cos^2 	heta)$$6 \cos^3 	heta + \cos^2 	heta - 1 = 0$Let $x = \cos 	heta$. Then $6x^3 + x^2 - 1 = 0$.By inspection,$x = \frac{1}{2}$ is a root: $6(\frac{1}{8}) + \frac{1}{4} - 1 = \frac{3}{4} + \frac{1}{4} - 1 = 0$.So,$\cos 	heta = \frac{1}{2} = \cos(\frac{\pi}{3})$.The general solution is $	heta = 2n\pi \pm \frac{\pi}{3}, n \in Z$.

If $\frac{1}{6} \sin \theta, \cos \theta, \tan \theta$ are in $G.P.$,then the general solution of $\theta$ is

Similar Questions

If $\alpha = 22^\circ 30'$,then $(1 + \cos \alpha )(1 + \cos 3\alpha )(1 + \cos 5\alpha )(1 + \cos 7\alpha )$ equals

If $\sin \theta + \sin^2 \theta = 1$ and $\cos^{12} \theta + a \cos^{10} \theta + b \cos^8 \theta + c \cos^6 \theta + d = 0$,then:

For $0 \leq P, Q \leq \frac{\pi}{2},$ if $\sin P + \cos Q = 2,$ then the value of $\tan \left(\frac{P + Q}{2}\right)$ is equal to

If $\sin x \cosh y = \cos \theta$,$\cos x \sinh y = \sin \theta$ and $4 \tan x = 3$,then $\sinh^2 y =$

If $5(\tan^2 x - \cos^2 x) = 2\cos 2x + 9$,then $\cos 4x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module