5 [__I‘__j\zx+I e é

1.	5 [__I‘__j\zx+I e é
Answer» y= tan^-1[{√(1+x^2)-1}/x]. Putting x=cot(a) y= tan^-1[{√(1+cot^2a)-1}/cota}]. y= tan^-1[{ 1/sina -1}×sina/cosa}]. y = tan^-1[(1-sina)/cosa ] y=tan^-1[(cos^2a/2+sin^2a/2–2.sina/2.cosa/2)/(cos^2a/2-sin^a/2)] y= tan^-1[(cosa/2-sina/2)^2/(cosa/2+sina/2)(cosa/2-sina/2)] y= tan^-1[(cosa/2-sina/2)/(cosa/2+sina/2)] Dividing in Nr and Dr by cosa/2 y= tan^-1[(1-tana/2)/(1+tana/2)] y= tan^-1[(tanπ/4-tana/2)/(1+tanπ/4.tanA/2)] y=tan^-1 tan(π/4-a/2) y = π/4 -a/2 y= π/4 -1/2.cot^-1(x) dy/dx = 0 -1/2.[-1/(1+x^2)]. dy/dx = 1/{2(1+x^2)}. Answer.

5 [I‘j\zx+I e é

Answer»

y= tan^-1[{√(1+x^2)-1}/x].

Putting x=cot(a)

y= tan^-1[{√(1+cot^2a)-1}/cota}].

y= tan^-1[{ 1/sina -1}×sina/cosa}].

y = tan^-1[(1-sina)/cosa ]

y=tan^-1[(cos^2a/2+sin^2a/2–2.sina/2.cosa/2)/(cos^2a/2-sin^a/2)]

y= tan^-1[(cosa/2-sina/2)^2/(cosa/2+sina/2)(cosa/2-sina/2)]

y= tan^-1[(cosa/2-sina/2)/(cosa/2+sina/2)]

Dividing in Nr and Dr by cosa/2

y= tan^-1[(1-tana/2)/(1+tana/2)]

y= tan^-1[(tanπ/4-tana/2)/(1+tanπ/4.tanA/2)]

y=tan^-1 tan(π/4-a/2)

y = π/4 -a/2

y= π/4 -1/2.cot^-1(x)

dy/dx = 0 -1/2.[-1/(1+x^2)].

dy/dx = 1/{2(1+x^2)}. Answer.