Question:
What is the value of $6 \tan ^{2} \theta-\frac{6}{\cos ^{2} \theta}$.
Solution:
We have,
$6 \tan ^{2} \theta-\frac{6}{\cos ^{2} \theta}=6 \tan ^{2} \theta-6 \sec ^{2} \theta$
$=-6\left(\sec ^{2} \theta-\tan ^{2} \theta\right)$
We know that, $\sec ^{2} \theta-\tan ^{2} \theta=1$
Therefore, $6 \tan ^{2} \theta-\frac{6}{\cos ^{2} \theta}=-6$
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