Question

g) $\sin^2 y \tan^2 y = \tan^2 y - \sin^2 y$ h) $\frac{\cos \theta}{1 + \sin \theta} + \tan \theta = \sec \theta$

          g) $\sin^2 y \tan^2 y = \tan^2 y - \sin^2 y$

h) $\frac{\cos \theta}{1 + \sin \theta} + \tan \theta = \sec \theta$

g) sin^2 y tan^2 y = tan^2 y - sin^2 y

h) (cosθ)/(1 + sinθ) + tanθ = secθ

Added by Andrew P.

Calculus, Early Transcendentals

Michael Sullivan, Kathleen Miranda 1st Edition

Chapter 2

Instant Answer

Solved by Expert Federico Castro

Step 1

1) tan y = tan y - sin y FALL/SUMM cos h We can simplify this equation by adding sin y to both sides: tan y + sin y = tan y - sin y + sin y tan y + sin y = tan y Since the left side and right side are equal, we can conclude that this equation is always true. Show more…

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