Prove that , (sin a -cos a +1)/(sin a + cos a -1) = 1/sec a - tan a
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Answers
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)
Proof
Taking L.H.S.
⇒ (sinA - cos a +1)/(sinA + cosA -1)
Divide the numerator and denominator by 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
||✪✪ 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,