Q55 / 120 · VITEEE 2006 · Mathematics
If $\sin \theta, \cos \theta, \tan \theta$ are in G.P. then $\cos ^{9} \theta+$ $\cos ^{6} \theta+3 \cos ^{5} \theta-1$ is equal to
- A$-1$
- Correct answerB0
- C1
- D2
Solution
Given : $\sin \theta, \cos \theta, \tan \theta$ are in G.P.
$\begin{array}{l}
\Rightarrow \cos ^{2} \theta=\sin \theta \tan \theta \Rightarrow \cos ^{3} \theta=\sin ^{2} \theta \\
\Rightarrow \cos ^{3} \theta=1-\cos ^{2} \theta \\
\Rightarrow\left(\cos ^{3} \theta+\cos ^{2} \theta\right)=1 ...(1)
\end{array}$
Cubic both sides, we have
$\begin{array}{c}
\cos ^{9} \theta+\cos ^{6} \theta+3 \cos ^{5} \theta \cdot\left(\cos ^{3} \theta+\cos ^{2} \theta\right)=1 \\
\Rightarrow \cos ^{9} \theta+\cos ^{6} \theta+3 \cos ^{5} \theta=1 \\
\text { [Using equation (1)] } \\
\Rightarrow \cos ^{9} \theta+\cos ^{6} \theta+3 \cos ^{5} \theta-1=0
\end{array}$
Hence, the correct answer is option: 'B'.