If $\tan A = \cot B$,prove that $A + B = 90^{\circ}$.
(N/A) Given that,$\tan A = \cot B$.
We know that $\cot B = \tan(90^{\circ} - B)$.
Substituting this into the given equation,we get $\tan A = \tan(90^{\circ} - B)$.
Since the tangent functions are equal,their angles must be equal: $A = 90^{\circ} - B$.
Rearranging the terms,we get $A + B = 90^{\circ}$.
We know that $\cot B = \tan(90^{\circ} - B)$.
Substituting this into the given equation,we get $\tan A = \tan(90^{\circ} - B)$.
Since the tangent functions are equal,their angles must be equal: $A = 90^{\circ} - B$.
Rearranging the terms,we get $A + B = 90^{\circ}$.
Explore More
Similar Questions
If $\cot \theta = \frac{7}{8},$ evaluate:
$(i) \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
$(ii) \cot^2 \theta$
$(i) \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
$(ii) \cot^2 \theta$
Difficult
View SolutionState whether the following is true or false. Justify your answer.
$\cot A$ is not defined for $A = 0^{\circ}$.
$\cot A$ is not defined for $A = 0^{\circ}$.
Easy
View SolutionIf $\tan 2A = \cot(A - 18^{\circ})$,where $2A$ is an acute angle,find the value of $A$ (in $^{\circ}$).
Medium
View SolutionState whether the following is true or false. Justify your answer.
The value of $\cos \theta$ increases as $\theta$ increases.
The value of $\cos \theta$ increases as $\theta$ increases.
Easy
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 SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo