The value of sec ^{2} 12^{circ}-frac{1}{tan ^ | vedclass

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Question

(B) Given expression: $\sec ^{2} 12^{\circ}-\frac{1}{	an ^{2} 78^{\circ}}$Since $\frac{1}{	an 	heta} = \cot 	heta$,the expression becomes $\sec ^{

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given expression: $\sec ^{2} 12^{\circ}-\frac{1}{	an ^{2} 78^{\circ}}$Since $\frac{1}{	an 	heta} = \cot 	heta$,the expression becomes $\sec ^{2} 12^{\circ}-\cot ^{2} 78^{\circ}$.Using the complementary angle identity $\cot(90^{\circ} - 	heta) = 	an 	heta$,we have $\cot 78^{\circ} = \cot(90^{\circ} - 12^{\circ}) = 	an 12^{\circ}$.Substituting this into the expression: $\sec ^{2} 12^{\circ}-	an ^{2} 12^{\circ}$.Using the trigonometric identity $\sec ^{2} 	heta - 	an ^{2} 	heta = 1$,the value is $1$.

The value of $\sec ^{2} 12^{\circ}-\frac{1}{\tan ^{2} 78^{\circ}}$ is

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