#(tan^3A-1)/(tanA-1)=sec^2A+tanA# prove?

1 Answer
Ratnaker Mehta
May 25, 2018

Please refer to a Proof in the Explanation.

Explanation:

I hope, the Question is to prove :

#(tan^3A-1)/(tanA-1)=sec^2A+tanA#.

We have, #(tan^3A-1)/(tanA-1)#,

#={cancel((tanA-1))(tan^2A+tanA+1)}/cancel((tanA-1))#,

#=(tan^2A+1)+tanA#,

#=sec^2A+tanA#, as desired!

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