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Verify the identity by transforming the left-hand side into the right-hand side.


(1 + cos² 3θ) / (sin² 3θ) = 2 csc² 3θ - 1

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olemakpadu

(1 + cos² 3θ) / (sin² 3θ) = 2 csc² 3θ - 1


Starting with the left:  Note that  cos²θ + sin²θ = 1.
In the same way:    cos²3θ + sin²3θ = 1
Therefore                cos²3θ  = 1 - sin²3θ
 From the top:  (1 + cos² 3θ) = 1 + 1 - sin²3θ = 2 - sin²3θ

(1 + cos² 3θ) / (sin² 3θ) = (2 - sin²3θ) / (sin² 3θ) = 2/sin² 3θ - sin²3θ/sin²3θ

=  2/sin² 3θ - 1;        But 1/sinθ = cscθ,  Similarly  1/sin3θ = csc3θ

= 2 *(1/sin3θ)²  - 1
= 2csc²3θ  - 1.        Therefore  LHS = RHS.  QED.

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