$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=$
- A$\tan 90^{\circ}$
- B$1$
- C$0$
- D$\sin 45^{\circ}$
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Similar Questions
Prove the following identity,where the angles involved are acute angles for which the expressions are defined:
$\frac{1+\sec A}{\sec A} = \frac{\sin^2 A}{1-\cos A}$
$\frac{1+\sec A}{\sec A} = \frac{\sin^2 A}{1-\cos A}$
Medium
View SolutionEvaluate:
$\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}$
$\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}$
Medium
View SolutionIn a right triangle $ABC$,right-angled at $B$. If $\tan A = 1$,then verify that $2 \sin A \cos A = 1$.
Easy
View SolutionEvaluate the following:
$\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$
$\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$
Medium
View SolutionShow that:
$(i)$ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} = 1$
$(ii)$ $\cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = 0$
$(i)$ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} = 1$
$(ii)$ $\cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = 0$
Medium
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