Prove that


$\left(\sin ^{2} 30^{\circ}-\sec ^{2} 60^{\circ}+4 \cot ^{2} 45^{\circ}\right)=?$

(a) 4
(b) 2
(c) 1

(d) $\frac{1}{4}$



As we know that,

$\sin 30^{\circ}=\frac{1}{2}$

$\sec 60^{\circ}=2$


$\cot 45^{\circ}=1$

By substituting these values, we get

$\left(\sin ^{2} 30^{\circ}-\sec ^{2} 60^{\circ}+4 \cot ^{2} 45^{\circ}\right)=\left(\frac{1}{2}\right)^{2}-(2)^{2}+4(1)^{2}$



Hence, the correct option is (d).


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