Question : What is the value of $\frac{3}{5}(\sec^2 20^\circ - \cot^2 70^\circ)$?
Option 1: $\frac{4}{3}$
Option 2: $\frac{5}{3}$
Option 3: $\frac{2}{5}$
Option 4: $\frac{3}{5}$
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Correct Answer: $\frac{3}{5}$
Solution :
$\frac{3}{5}(\sec^2 20^\circ - \cot^2 70^\circ)$
$=\frac{3}{5}(\sec^2 20^\circ - \tan^2 20^\circ)$ [By using: $\tan\theta = \cot(90^\circ-\theta)$]
$=\frac{3}{5}\times1$ [We know that $\sec^2 \theta- \tan^2 \theta = 1$]
$=\frac{3}{5}$
Hence, the correct answer is $\frac{3}{5}$.
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