Question

Cos2a/1-tana+sin3a/sina-cosa=1+sina.cosa

Answer 1

✨✨heymate here your soln ✨✨


Prove : cos2A1−tanA+sin3AsinA−cosA=1+sinAcosA.

We have, cos2A1−tanA+sin3AsinA−cosA,

=cos2A1−sinAcosA−sin3AcosA−sinA,

=cos2AcosA−sinAcosA−sin3AcosA−sinA,

=cos3AcosA−sinA−sin3AcosA−sinA,

=cos3A−sin3AcosA−sinA,

=(cosA−sinA)(cos2A+cosAsinA+sin2A)(cosA−sinA)..........(∗),

=(cos2A+sin2A+sinAcosA),

=1+sinAcosA, as desired!

Note that while cancelling (cosA−sinA) at (∗), we have

assumed that (cosA−sinA)≠0,i.e.,tanA≠1.

This is quite admissible, otherwise the left member of the

Identity would not exist!

Answer 2

Answer:

Hyyy dear friend your answer....

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