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tan X/2 = t ,sinx = 2t/1+t2 ; cos x = 1–t2 / 1+t2, x = 2 tan–1t; dx = 2dt/1+t2
how we derive dx = 2dt/1+t2
thanks
You can differentiate any of the above equations with respect to t.
take
sinx = 2t/1+t2
cosxdx = ((1+t^2)2 - 4t^2)/(1+t^2)^2 * dt
1-t^2/1+t^2 *dx = 2* (1-t^2)/(1+t^2)^2 * dt
cancelling on Lhs and Rhs
dx = 2dt/1+t^2
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