(C) Given that $\tan A = \cot B$.We know the trigonometric identity $\cot B = \tan(90^\circ - B)$.Substituting this into the equation,we get $\tan A =

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(C) Given that $\tan A = \cot B$.We know the trigonometric identity $\cot B = \tan(90^\circ - B)$.Substituting this into the equation,we get $\tan A = \tan(90^\circ - B)$.By comparing the angles,we have $A = 90^\circ - B$.Rearranging the terms,we get $A + B = 90^\circ$.

Class 10 Mathematics Introduction to Trigonometry Mix Examples - Introduction to Trigonometry

Easy

If $\tan A = \cot B$,then $A + B = \ldots$

A
$30$
B
$60$
C
$90$
D
$180$

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Introduction to Trigonometry

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Mix Examples - Introduction to Trigonometry

Given that $\sin \alpha = \frac{1}{2}$ and $\cos \beta = \frac{1}{2}$,then the value of $(\alpha + \beta)$ is (in $^{\circ}$)

Easy

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Prove that $\frac{1+\sec \theta-\tan \theta}{1+\sec \theta+\tan \theta}=\frac{1-\sin \theta}{\cos \theta}$

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Prove that $\sin^{6} \theta + \cos^{6} \theta + 3 \sin^{2} \theta \cos^{2} \theta = 1$.

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If $\tan \theta = \sqrt{3}$,then $\theta = \ldots$ (in $^\circ$)

Easy

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$\sin ^{2} 1^{\circ} + \sin ^{2} 3^{\circ} + \sin ^{2} 87^{\circ} + \sin ^{2} 89^{\circ} = \ldots$

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If $\tan A = \cot B$,then $A + B = \ldots$

Similar Questions

Given that $\sin \alpha = \frac{1}{2}$ and $\cos \beta = \frac{1}{2}$,then the value of $(\alpha + \beta)$ is (in $^{\circ}$)

Prove that $\frac{1+\sec \theta-\tan \theta}{1+\sec \theta+\tan \theta}=\frac{1-\sin \theta}{\cos \theta}$

Prove that $\sin^{6} \theta + \cos^{6} \theta + 3 \sin^{2} \theta \cos^{2} \theta = 1$.

If $\tan \theta = \sqrt{3}$,then $\theta = \ldots$ (in $^\circ$)

$\sin ^{2} 1^{\circ} + \sin ^{2} 3^{\circ} + \sin ^{2} 87^{\circ} + \sin ^{2} 89^{\circ} = \ldots$

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