# Is it right: 2/sinx - 2sinx cosx /1 = 2-2sin^2 cosx / sinx = 2- 2 sinxcosx = 2(1-sinx cosx) = 2(cos^2 - sinxcosx + sin^2) = 2*1 = 2 ?

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• Is it right: 2/sinx - 2sinx cosx /1 = 2-2sin^2 cosx / sinx = 2- 2 sinxcosx = 2(1-sinx cosx) = 2(cos^2 - sinxcosx + sin^2) = 2*1 = 2 ?

Is it right: 2/sinx - 2sinx cosx /1 = 2-2sin^2 cosx / sinx = 2- 2 sinxcosx = 2(1-sinx cosx) = 2(cos^2 - sinxcosx + sin^2) = 2*1 = 2 ? Find answers now! No ...
Positive: 63 %
Is it right: 2/sinx - 2sinx cosx /1 = 2-2sin^2 cosx / sinx = 2- 2 sinxcosx = 2(1-sinx cosx) = 2(cos^2 - sinxcosx + sin^2) = 2*1 = 2 ?
Positive: 60 %

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... such as the Pythagorean Theorem's "a 2 + b 2 = c 2" for right triangles. There are loads of trigonometric identities, ... sin 2 (t) + cos 2 (t) = 1 tan ...
Positive: 63 %
Using the identity cos 2 + sin = 1 we have sin2 1 ... Starting from the right-hand side to obtain 1 1 sinx 1 ... 1 sin2 x = 2sinx cos2 x = 2 sinx cosx 1 cosx
Positive: 58 %
... (sinx-cosx)(sinx+cosx) 2.(1+tanx)^2 3.(1-sinx) ... *cos(x) - cos^2(x) reducing sin^2(x) - cos^2(x) 2) (1+ tan(x))^2 ... (1-cosx)^1/2*(1+cosx)^1/2=sinx;