Question

Prove that , (sin a -cos a +1)/(sin a + cos a -1) = 1/sec a - tan a
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Answer 1

Given

(sinA - cos a +1)/(sinA + cosA -1) = 1/(secA - tanA)

To prove

Prove that, (sinA - cos a +1)/(sinA + cosA -1) = 1/(secA - tanA)

$\rule{200}2$

Proof

Taking L.H.S.

⇒ (sinA - cos a +1)/(sinA + cosA -1)

Divide the numerator and denominator by cosA

⇒ $\dfrac{\frac{sinA}{cosA} -\frac{cosA}{cosA}+\frac{1}{cosA}}{ \frac{sinA}{cosA} +\frac{cosA}{cosA}-\frac{1}{cosA}}$

We know that, 1/cosA = secA

⇒ (tanA - 1 + secA)/(tanA + 1 - secA)

Also, sec²A - tan²A = 1

⇒ [tanA + secA - 1]/[tanA - secA + (sec²A - tan²A)]

Used identity: (a² - b²)= (a - b) (a + b)

⇒ (tanA + secA - 1)/[tanA - secA + (secA - tanA)(secA + tanA)]

Take (secA - tanA) as common

⇒ (tanA + secA - 1)/[(secA - tanA)(tanA + secA - 1)

(tanA + secA - 1) cancel out, we left with

⇒ 1/(secA - tanA)

L.H.S. = R.H.S.

Hence, proved

Answer 2

||✪✪ QUESTION ✪✪||

Prove that , (sinA -cosA +1)/(sinA + cosA -1) = 1/(secA - tanA)

|| ✰✰ ANSWER ✰✰ ||

Taking LHS,

→ (sinA -cosA +1)/(sinA + cosA -1)

Dividing both Numerator and Denominator by cosA we get,

→ [(sinA/cosA) - (cosA/cosA) + (1/cosA)] / [ (sinA/cosA) + (cosA/cosA) - (1/cosA)]

Now, Putting (sinA/cosA) = TanA , and (1/cosA) = secA ,

→ [ (TanA - 1 + secA ) / (TanA + 1 - secA) ]

→ [ (TanA + secA - 1) / (TanA - secA + 1) ]

Now, Putting 1 = sec²A - tan²A in The Denominator , we get,

→ [ (TanA + secA - 1) / (TanA - secA + (sec²A - tan²A) ]

Now, using (a² - b²) = (a+b)(a-b) in The Denominator,

→ [ (TanA + secA - 1) / (TanA - secA + {(secA - tanA)(SecA + TanA)} ]

Taking (-1) common From The denominator now,

→ [ (TanA + secA - 1)] /[{(-1)*(secA - TanA)} + {(secA - tanA)(SecA + TanA)} ]

Now, Taking (secA - tanA) common From Denominator ,

→ [ (TanA + secA - 1)] / [ (secA -TanA){ (-1) + secA + TanA) ]

→ [ (TanA + secA - 1)] / [(secA - TanA)(TanA + secA - 1)]

(TanA + secA - 1) will be cancel from both sides now,

we get,

→ 1/(seA - TanA) = RHS.

✪✪ Hence Proved ✪✪