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  • 19-11-2018
  • Mathematics
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((1-cos^2)/tan^2)+2sin^2

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gmany
gmany gmany
  • 25-11-2018

[tex]\dfrac{1-\cos^2x}{\tan^2x}+2\sin^2x\\\\\text{use}\ \tan x=\dfrac{\sin x}{\cos x}\ \text{and}\ \sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x\\\\=\dfrac{\sin^2x}{\left(\frac{\sin x}{\cos x}\right)^2}+2\sin^2x=\dfrac{\sin^2x}{\frac{\sin^2x}{\cos^2x}}+2\sin^2x=\sin^2x\cdot\dfrac{\cos^2x}{\sin^2x}+2\sin^2x\\\\=\cos^2x+2\sin^2x=\cos^2x+2(1-\cos^2x)\\\\\text{use distributive property}\\\\=\cos^2x+2-2\cos^2x=2-\cos^2x[/tex]

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