Question : What is the value of $3 \sin ^2 30^{\circ}+\frac{3}{5} \cos ^2 60^{\circ}–2 \sec ^2 45^{\circ} $?
Option 1: $-\frac{5}{2}$
Option 2: $-\frac{5}{8}$
Option 3: $-\frac{31}{10}$
Option 4: $-\frac{25}{17}$
Correct Answer: $-\frac{31}{10}$
Solution :
Given: The trigonometric expression is $3 \sin ^2 30^{\circ}+\frac{3}{5} \cos ^2 60^{\circ}–2 \sec ^2 45^{\circ} $.
We know the values of trigonometric ratios, $\sin 30^{\circ} =\frac{1}{2}$, $\cos 60^{\circ}=\frac{1}{2}$ and $\sec 45^{\circ}=\sqrt2$.
⇒ $3 \sin ^2 30^{\circ}+\frac{3}{5} \cos ^2 60^{\circ}–2 \sec ^2 45^{\circ}$
$= 3\times(\frac{1}{2})^2+\frac{3}{5}\times(\frac{1}{2})^2–2 \times (\sqrt2)^2 $
$= \frac{3}{4}+\frac{3}{20}-4=\frac{15+3-80}{20}$
$= \frac{18-80}{20}=-\frac{62}{20}=-\frac{31}{10}$
Hence, the correct answer is $-\frac{31}{10}$.
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