MCQ
$\frac{3}{4} \tan ^2 30^{\circ}-\sec ^2 45^{\circ}+\sin ^2 60^{\circ}$ is equal to
- ✓-1
- B$\frac{5}{6}$
- C$-\frac{3}{2}$
- D$\frac{1}{6}$
✓
Answer
Correct option: A.
-1
(A)-1
$
\begin{aligned}
\text { } & \frac{3}{4} \tan ^2 30^{\circ}-\sec ^2 45^{\circ}+\sin ^2 60^{\circ} \\
= & \frac{3}{4}\left(\frac{1}{\sqrt{3}}\right)^2-(\sqrt{2})^2+\left(\frac{\sqrt{3}}{2}\right)^2=\frac{3}{4} \times \frac{1}{3}-2+\frac{3}{4}=1-2=-1
\end{aligned}
$
$
\begin{aligned}
\text { } & \frac{3}{4} \tan ^2 30^{\circ}-\sec ^2 45^{\circ}+\sin ^2 60^{\circ} \\
= & \frac{3}{4}\left(\frac{1}{\sqrt{3}}\right)^2-(\sqrt{2})^2+\left(\frac{\sqrt{3}}{2}\right)^2=\frac{3}{4} \times \frac{1}{3}-2+\frac{3}{4}=1-2=-1
\end{aligned}
$
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