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tan X/2 = t ,sinx = 2t/1+t 2 ; cos x = 1–t 2 / 1+t 2 , x = 2 tan –1 t; dx = 2dt/1+t 2 how we derive dx = 2dt/1+t 2 thanks

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


Grade:

1 Answers

Harish R
37 Points
8 years ago

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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