(C) We know that $\tan 45^{\circ} = 1$.Substituting this value into the expression:$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}} = \frac{1-(1

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(C) We know that $\tan 45^{\circ} = 1$.Substituting this value into the expression:$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}} = \frac{1-(1)^{2}}{1+(1)^{2}}$$= \frac{1-1}{1+1}$$= \frac{0}{2}$$= 0$Therefore,the correct option is $C$.

Class 10 Mathematics Introduction to Trigonometry Textbook - Introduction to Trigonometry

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$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=$

A
$\tan 90^{\circ}$
B
$1$
C
$0$
D
$\sin 45^{\circ}$

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Introduction to Trigonometry

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Textbook - Introduction to Trigonometry

Prove the following identity,where the angle involved is an acute angle for which the expression is defined:
$\sqrt{\frac{1+\sin A}{1-\sin A}} = \sec A + \tan A$

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$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}} = ?$

Medium

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Evaluate:
$\cos 48^{\circ}-\sin 42^{\circ}$

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If $\sin 3A = \cos(A - 26^{\circ})$,where $3A$ is an acute angle,find the value of $A$.

Medium

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$9 \sec^{2} A - 9 \tan^{2} A = \dots$

Easy

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$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=$

Similar Questions

Prove the following identity,where the angle involved is an acute angle for which the expression is defined:$\sqrt{\frac{1+\sin A}{1-\sin A}} = \sec A + \tan A$

$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}} = ?$

Evaluate:$\cos 48^{\circ}-\sin 42^{\circ}$

If $\sin 3A = \cos(A - 26^{\circ})$,where $3A$ is an acute angle,find the value of $A$.

$9 \sec^{2} A - 9 \tan^{2} A = \dots$

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Prove the following identity,where the angle involved is an acute angle for which the expression is defined:
$\sqrt{\frac{1+\sin A}{1-\sin A}} = \sec A + \tan A$

Evaluate:
$\cos 48^{\circ}-\sin 42^{\circ}$