Question:
$\left(\sin ^{2} 30^{\circ}-\sec ^{2} 60^{\circ}+4 \cot ^{2} 45^{\circ}\right)=?$
(a) 4
(b) 2
(c) 1
(d) $\frac{1}{4}$
Solution:
As we know that,
$\sin 30^{\circ}=\frac{1}{2}$
$\sec 60^{\circ}=2$
$\cot 45^{\circ}=1$
By substituting these values, we get
$\left(\sin ^{2} 30^{\circ}-\sec ^{2} 60^{\circ}+4 \cot ^{2} 45^{\circ}\right)=\left(\frac{1}{2}\right)^{2}-(2)^{2}+4(1)^{2}$
$=\frac{1}{4}-4+4$
$=\frac{1}{4}$
Hence, the correct option is (d).