Question

PLEASE ANSWER MY QUESTION GUYS

Answer 1

$Given :\frac{tanA + secA - 1}{tanA - secA + 1}$

We know that tan^2A + 1 = sec^2A (or) tan^2A - sec^2A = -1

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

We know that a^2 - b^2 = (a + b)(a - b)

$=>\frac{tanA + secA + (tanA - secA)(tanA + secA)}{tanA - secA + 1}$

Take tanA + secA as common, we get

$=> \frac{tanA + secA(tanA - secA + 1)}{tanA - secA + 1}$

$=> tanA + secA$

$=>\frac{sinA}{cosA}+\frac{1}{cosA}$

$=>\boxed{\frac{1 + sinA}{cosA}}$

Hope it helps!

Answer 2

Here is your answer:

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Given:

LHS

$\frac{tanA+secA-1}{tanA-secA+1}$

We know that : $sec^2A - tan^2A = 1$

$\implies \frac{tanA + secA - (sec^2A-tan^2A)}{tanA-secA+1}$

$\implies \frac{tanA + secA +tan^2A-sec^2A)}{tanA-secA+1}$

We know that $a^2 - b^2 = (a+b)(a-b)$

$\implies \frac{tanA+secA +(tanA+secA)(tanA-secA)}{tanA-secA+1}$

Take $tanA+ sec A$ as common:

$\implies \frac{(tanA + secA )( 1 + tan A - sec A)}{tanA - secA + 1}$

Cancel the term $tan A - secA + 1$

$\implies tanA + sec A$

$tan A = \frac{sinA}{cosA}$

$secA = \frac{1}{cosA}$

$\implies \frac{sinA}{cosA} + \frac{1}{cosA}$

$\implies \frac{ 1 + sinA}{cosA}$ ===> RHS

Thus LHS=RHS

Hence proved.

Hope it helps

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