Question

If sin $\theta$ , cos $\theta$ , tan $\theta$ are in G.P. then $cos^{9} \theta + cos^{6}\theta + 3cos^{5}\theta- 1$ is equal to

• A. -1
• B. 0
• C. 1
• D. 2

Answer 1

Answer: (B)

0

Answer 2

Given : sin $\theta$ , cos $\theta$ , tan $\theta$ are in G.P.
$\Rightarrow \, \, \, cos^2 \theta \, = \, sin\theta \, tan\theta \, \Rightarrow \, cos^3 \theta \, = \, sin^2\theta$
$\Rightarrow \, \, cos^3 \theta \, =1-cos^2 \theta$
$\Rightarrow \, \, \, (cos^3\theta \, \, + \, cos^2 \theta) \, = 1 \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ...(1)$
Cubic both sides, we have
$cos^9 \theta + cos^6 \theta + 3 cos^5 \theta . (cos^3 \theta+ cos^2 \theta) =1$
$\Rightarrow \, \, \, cos^9 \theta + cos^6 \theta + 3cos^5 \theta =1$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,$[Using equation (1)]
$\Rightarrow \, \, \, cos^9 \theta + cos^6 \theta + 3cos^5 \theta - 1=0$