Question

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

Answer 1

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....

Answer 2

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