Prove that this⤴️⤴️
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Answers
Solution
Prove That :-
- (sin A - cos A + 1)/(sin A + cos A - 1) = 1/(sec A - tan A )
Proof :-
Take L.H.S.
= (sin A - cos A + 1)/(sin A + cos A - 1)
Divide by cos A Numerator & Dominator
= ( sin A/cos A - cos A/cos A + 1/cos A)/(sinA/cos A + cos A/cos A - 1/cos A)
We know,
★ sin A/cos A = tan A
★ 1/cos A = sec A
So, Now
= ( tan A - 1 + sec A )/( tan A + 1 - sec A )
= { tan A + sec A) - ( sec² A - tan² A)}\(tan A + 1 - sec A )
= {( tan A + sec A )-(secA - tan A )(sec A + tan A)/(tanA + 1 - sec A)
= (tan A + sec A )(1 -sec A + tan A)/(tanA + 1 - sec A)
Divide by (tanA + 1 - sec A) Numerator & Denominator
= ( tan A + sec A)
Multiply by (sec A - tan A) Numerator & Denominator
= (tan A + sec A)(sec A - tan A)/(sec A - tan A)
= ( sec² A - tan² A )/( sec A - tan A )
We Know,
- ( sec² A - tan² A) = 1
= 1/(sec A - tan A )
= R.H.S.
Hence Proved
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