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N76 [4]
3 years ago
6

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

Mathematics
2 answers:
sammy [17]3 years ago
7 0

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

<h3>sec²A - 1 = tan ²A</h3>

So we have the final answer as

<h3>tan²A</h3>

As proven

Hope this helps you

Nady [450]3 years ago
3 0

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.

<em><u>Hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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