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If $\sin \theta, \cos \theta$ and $\tan \theta$ are in $\mathrm{GP}$, then $\cot ^{6} \theta-\cot ^{2} \theta$ is
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The correct answer is:
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$\because \sin \theta, \cos \theta$ and $\tan \theta$ are in GP.
$$
\begin{array}{ll}
\text { So, } & \cos ^{2} \theta=\sin \theta \cdot \tan \theta \\
\Rightarrow & \cos ^{3} \theta=\sin ^{2} \theta \\
\Rightarrow \quad & \cot ^{3} \theta=\frac{1}{\sin \theta}=\operatorname{cosec} \theta
\end{array}
$$
Squaring both sides, we get
$$
\begin{gathered}
\cot ^{6} \theta=\operatorname{cosec}^{2} \theta \\
\Rightarrow \quad \cot ^{6} \theta=1+\cot ^{2} \theta \\
\Rightarrow \quad \cot ^{6} \theta-\cot ^{2} \theta=1
\end{gathered}
$$
$$
\begin{array}{ll}
\text { So, } & \cos ^{2} \theta=\sin \theta \cdot \tan \theta \\
\Rightarrow & \cos ^{3} \theta=\sin ^{2} \theta \\
\Rightarrow \quad & \cot ^{3} \theta=\frac{1}{\sin \theta}=\operatorname{cosec} \theta
\end{array}
$$
Squaring both sides, we get
$$
\begin{gathered}
\cot ^{6} \theta=\operatorname{cosec}^{2} \theta \\
\Rightarrow \quad \cot ^{6} \theta=1+\cot ^{2} \theta \\
\Rightarrow \quad \cot ^{6} \theta-\cot ^{2} \theta=1
\end{gathered}
$$
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