Question

please solve it out friends it's too urgent please please please.....

Answer 1

Hi there!
Here's the answer:

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¶ Sec² x - tan² x = 1

¶ (a² - b²) = (a+b)(a-b)

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SOLUTION :

L.H.S =

$\frac{tan A + Sec A - 1}{tan A - Sec A +1}$

= $\frac{tan A + Sec A - (sec^{2}A - tan^{2} A)}{tan A - Sec A +1}$

= $\frac{tan A + Sec A - [(tan A + Sec A)(Sec A - tan A)]}{tan A - Sec A + 1}$

°•° a- ab = a(1-b)

Here a = tan A + sec A
b = tan A - sec A


= $\frac{(tan A + Sec A)( 1 - [sec A - tan A])}{tan A - Sec A + 1}$

= $\frac{(tan A + Sec A( 1 -sec A + tan A)}{tan A - Sec A + 1}$

= $\frac{(tan A + Sec A( 1 -sec A + tan A)}{1 - Sec A + tan A}$


=$tan A + Sec A$

=$\frac{Sin A}{CosA} + \frac{1}{Cos A}$

= $\frac{1 + Sin A}{Cos A}$

= R.H.S

Hence Proved.

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Hope it helps

Answer 2

here is your answer sis ❤❤❤❤❤

HOPE IT HELPS U
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