Question
What is the value of $6\tan^2\theta-\frac{6}{\cos^2\theta}?$
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Answer
$6\tan^2\theta-\frac{6}{\cos^2\theta}=6\tan^2-6\sec^2\theta$
$\begin{cases}\because \frac{1}{\cos\theta}=\sec\theta\end{cases}$
$=-6(\sec^2\theta-\tan^2\theta)\{\sec^2\theta-\tan^2\theta=1\}$
$=-6\times1=-6$
$\begin{cases}\because \frac{1}{\cos\theta}=\sec\theta\end{cases}$
$=-6(\sec^2\theta-\tan^2\theta)\{\sec^2\theta-\tan^2\theta=1\}$
$=-6\times1=-6$
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