Question

sinθ-cosθ+1/sinθ+cosθ-1=1/secθ-tanθ Prove it?

Answer 1

Proved

● L.H.S => secθ+tanθ
multiply by \times \frac{secθ - tanθ}{secθ - tanθ}

$= (secθ + tanθ) \times \frac{secθ - tanθ}{secθ - tanθ} \\ = ( {sec}^{2} θ - {tan}^{2} θ) \times \frac{1}{secθ - tanθ} \\ \frac{1}{secθ - tanθ} \\ proved$
L.H.S = R.H.S

■I HOPE ITS HELP■

Answer 2

Here, I am writing theta as A.

$Given : \frac{sinA - cosA + 1}{sinA + cosA - 1}$

Multiply by cosA.

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

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

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

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

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

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

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

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

$= > secA + tanA$

On rationalizing we get,

$= > secA + tanA * \frac{secA - tanA}{secA - tanA}$

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

$= > \frac{1}{secA - tanA}$

Hope it helps!