Question

Solve this problem using the Trigonometric identities (secA+1)(SecA-1)= tan^2A

Answer 1

Step-by-step explanation:

( secA + 1)( sec A - 1)

Using the expansion

( a + b)( a - b) = a² - b²

Expand the expression

We have

sec²A + secA - secA - 1

That's

sec² A - 1

From trigonometric identities

sec²A - 1 = tan ²A

So we have the final answer as

tan²A

As proven

Hope this helps you

Answer 2

Step-by-step explanation:

Here,

LHS

= (SecA+1)(secA -1)

$= {sec}^{2} A - 1$

${as{a}^{2} - {b}^{2} =(a + b)(a - b)$

Now, we have formula that:

${sec}^{2} \alpha - {tan \alpha }^{2} = 1$

${tan}^{2} \alpha = {sec }^{2} \alpha - 1$

as we got ,

$= {sec}^{2} A- 1$

This is equal to:

$= {tan}^{2} A$

= RHS proved.

Hope it helps....