harveenxxow6p2l harveenxxow6p2l
  • 12-09-2017
  • Mathematics
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Show that sinxtanx=1-cos^2x/cosx

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kettboy kettboy
  • 12-09-2017
LHS: = [tex]\sin x \tan x = \frac{\sin x \sin x }{\cos x} = \frac{\sin^2 x}{\cos x} [/tex] (using [tex]\tan x = \frac{\sin x}{\cos x} [/tex])

We know [tex]\sin^2x + \cos^2x = 1 \Rightarrow \sin^2x=1-\cos^2x[/tex] so we can replace the sin²x in the LHS expression as follows

[tex] \frac{\sin^2x}{\cos x} = \frac{1-\cos^2x}{\cos x} [/tex] which is the RHS.
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