How to prove 1/(1⁡+⁡sinx) =⁡ [(secx⁡-tanx)^2]/(1⁡-⁡sin⁡x)?

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Trigonometric Identities Part 3. Start solving the right-hand side members of the given equation, [(secx-tanx)^2]/(1-sinx), until the left-hand side members of the given equation, 1/(1+sinx), are obtained. The right-hand side members of the given equation are now {[(1/cosx)-(sinx/cosx)]^2}/(1-sinx) after replacing secx by 1/cosx, and tanx by sinx/cosx, a complex fraction with cosx and sinx terms. Evaluate and simplify.

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