Trig identity

sin B(sec B)^2 =27 cosB (cosecB)^2

How would you go about rearranging this in terms of tanB? What identities do you use?

Original post by stolenuniverse

sin B(sec B)^2 =27 cosB (cosecB)^2

How would you go about rearranging this in terms of tanB? What identities do you use?

Divide both sides by

\cos(B)\csc^2 (B)

and you should have

\tan^3(B)

on the LHS after some very little manipulation.

Try and get that, there are no other identities involved apart from

\tan(B)

in terms of sine and cosine.

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