Question

Prove the following trigonometric identities:
(secA+tanA-1)(secA-tanA+1)=2tanA

Answer 1

$(\sec A+\tan A-1)(\sec A-\tan A+1)=2\tan A$, proved.

Step-by-step explanation:

Prove that the identity:

$(\sec A+\tan A-1)(\sec A-\tan A+1)=2\tan A$.

L.H.S. = $(\sec A+\tan A-1)(\sec A-\tan A+1)$

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

= $\sec^2 A-\sec A\tan A+\sec A+\tan A\sec A-\tan^2 A+\tan A-\sec A+\tan A-1$

= $\sec^2 A-\tan^2 A+\tan A+\tan A-1$

Using the trigonometric identity,

$\sec^2 A-\tan^2 A$ = 1

= $1+\tan A+\tan A-1$

= $2\tan A$

= R.H.S., proved.

Thus, $(\sec A+\tan A-1)(\sec A-\tan A+1)=2\tan A$, proved.

Answer 2

(sec A + tan A −1 )(sec A − tan A + 1)=2tan A , proved.

[Prove that the identity:

$(\sec A+\tan A-1)(\sec A-\tan A+1)=2\tan A.$

$L.H.S. = (\sec A+\tan A-1)(\sec A-\tan$

$\implies \sec A(\sec A-\tan A+1)+\tan A(\sec A-\tan A+1)-1(\sec A-\tan A+1)$

$\implies \sec^2 A-\sec A\tan A+\sec A+\tan A\sec A-\tan^2 A+\tan A-\sec A+\tan A-1s$

$\implies \sec^2 A-\tan^2 A+\tan A+\tan A-$

Using the trigonometric identity,

$\sec^2 A-\tan^2 Asec$

$\implies 1+\tan A+\tan A-11+tanA+tanA−1$

$\implies 2\tan A2tanA$

= R.H.S., proved.

Thus,$(\sec A+\tan A-1)(\sec A-\tan A+1)=2\tan A, proved.$