Question

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SOLVE IT GUYS PLS . I WILL MARK BRAINIEST .

Answer 1

[Hello]
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Using Formula

Tan∅=Sin∅/Cos∅

Sec∅=1/Cos∅

Thanks!!!

Answer 2

(4)

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

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

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

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

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

$= > secA + tanA$

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

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

On rationalizing, we get

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

$= > \frac{(1 + sinA)(cosA)}{cos^2A}$

$= > \frac{(1 + sinA)(cosA)}{1 - sin^2A}$

$= > \frac{(1 + sinA)(cosA)}{(1 + sinA)(1 - sinA)}$

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

Hope it helps!