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Mathematics 12 Online
OpenStudy (anonymous):

True or false? (sinx+cotx/cosx)^2=tan^2x+2secx+csc^2x

OpenStudy (anonymous):

\[(sinx + cotx / cosx)^{2} = \tan ^{2}x + 2secx + \csc ^{2}x\]

OpenStudy (anonymous):

@Mertsj

OpenStudy (mertsj):

I would start by replacing cot with cos/sin then simplify the complex fraction and square. See where that leads.

OpenStudy (mertsj):

I'll do it too. Hopefully we will come to the same conclusion.

OpenStudy (anonymous):

when i square it do i square everything.. like numerator and denominator and thanks

OpenStudy (mertsj):

yes

OpenStudy (anonymous):

so i got: sin^2x + (cos^2x/sin^2x) / cos^2 = tan^2x + 2 (1/cosx) + 1/sin^2

OpenStudy (mertsj):

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