Math, asked by yamin1971mya, 4 months ago

Prove that this⤴️⤴️
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Answered by BrainlyYuVa
67

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

_________________

Answered by Anonymous
2

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