Prove that $\sec A(1-\sin A)(\sec A+\tan A)=1$.
(A) $LHS = \sec A(1-\sin A)(\sec A+\tan A)$
$= \left(\frac{1}{\cos A}\right)(1-\sin A)\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)$
$= \left(\frac{1-\sin A}{\cos A}\right)\left(\frac{1+\sin A}{\cos A}\right)$
$= \frac{(1-\sin A)(1+\sin A)}{\cos^2 A}$
$= \frac{1-\sin^2 A}{\cos^2 A}$
$= \frac{\cos^2 A}{\cos^2 A} = 1 = RHS$
Hence,the identity is proved.
$= \left(\frac{1}{\cos A}\right)(1-\sin A)\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)$
$= \left(\frac{1-\sin A}{\cos A}\right)\left(\frac{1+\sin A}{\cos A}\right)$
$= \frac{(1-\sin A)(1+\sin A)}{\cos^2 A}$
$= \frac{1-\sin^2 A}{\cos^2 A}$
$= \frac{\cos^2 A}{\cos^2 A} = 1 = RHS$
Hence,the identity is proved.
Explore More
Similar Questions
Prove the following identity,where the angle involved is an acute angle for which the expression is defined:
$\sqrt{\frac{1+\sin A}{1-\sin A}} = \sec A + \tan A$
$\sqrt{\frac{1+\sin A}{1-\sin A}} = \sec A + \tan A$
Medium
View SolutionProve the following identity,where the angles involved are acute angles for which the expressions are defined:
$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$
$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$
Difficult
View SolutionIf $\sin 3A = \cos(A - 26^{\circ})$,where $3A$ is an acute angle,find the value of $A$.
Medium
View Solution$\sin 2A = 2 \sin A$ is true when $A =$ (in $^{\circ}$)
Easy
View SolutionEvaluate the following:
$2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$
$2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$
Easy
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