Prove that: 1+sinAcosA+tanA=secA

L.H.S. =1+sinAcosA+tanA=1+sinAcosA+cosAsinA=(1+sinA)cosAcos2A+sinA(1+sinA)=(1+sinA)(cosA)cos2A+sinA+sin2A=(1+sinA)cosA1+sinA=cosA1=secA= R.H.S.

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Prove that: { A}{1 + A} + A = A

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Prove that: $\frac{cos A}{1 + sin A} + tan A = sec A$

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Prove that:

\frac{cos A}{1 + sin A} + tan A = sec A

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L.H.S.

= \frac{cos A}{1 + sin A} + tan A

= \frac{cos A}{1 + sin A} + \frac{sin A}{cos A}

= \frac{cos ^{2} A + sin A ( 1 + sin A )}{( 1 + sin A ) cos A}

= \frac{cos ^{2} A + sin A + sin ^{2} A}{( 1 + sin A ) ( cos A )}

= \frac{1 + sin A}{( 1 + sin A ) cos A}

= \frac{1}{cos A}

= sec A =

R.H.S.

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Question Text	Prove that: $\frac{cos A}{1 + sin A} + tan A = sec A$
Answer Type	Text solution:1
Upvotes	150