Question

sinAtanA/1-cosA=1+secA​

Answer 1

Refer to the attachment ✌

Hope it will helps ✌

$:)$

Answer 2

Answer:

Hi,

▶(sin A-cos A+1)/(sin A+cosA-1)=1/(sec A-tan A)

L.H.S. divide above and below by cos A

=(tan A-1+secA)/(tan A+1-sec A)

=(tan A-1+secA)/(1-sec A+tan A)

We know that 1+tan^2A=sec ^2A

Or 1=sec^2A-tan ^2A=(sec A+tan A)(secA-tanA)

=(sec A+tan A-1)/[(sec A+tan A)(sec A-tan A)-(sec A-tan A)]

=(sec A+tan A-1)/(sec A-tan A)(sec A+tan A-1)

= 1/(sec A-tan A) , proved.

Things are not looking good! At this stage, both sides should be seen to be equal!

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