Prove the following trigonometric identities:$ \sec A(1-\sin A)(\sec A+\tan A)=1 $

To do:

We have to prove that \( \sec A(1-\sin A)(\sec A+\tan A)=1 \).

Solution:

We know that,

$\tan A=\frac{\sin A}{\cos A}=\sin A\sec A$.....(i)

$\sec^2 A-tan ^{2} A=1$.......(ii)

Therefore,

$\sec A(1-\sin A)(\sec A+\tan A)=(\sec A-\sec A\sin A)(\sec A+\tan A)$ 

$=(\sec A-\tan A)(\sec A+\tan A)$           (From (i))

$=\sec^2 A-tan ^{2} A$                  [$(a+b)(a-b)=a^2-b^2$]

$=1$                          (From (ii))

Hence proved.    

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

40 Views

Kickstart Your Career

Get certified by completing the course

Get Started