Question

Find the derivative using substitution method
${tan}^{ - 1} \sqrt{ \frac{1 - t}{1 + t } }$
Please do it if you can, please do it fast...............
Best answer will be made brainliest................ ​

Answer 1

Answer:

$</p><p>{tan}^{ - 1} \sqrt{ \frac{1 - t}{1 + t } }$

$\sf Let~{tan}^{ - 1} \sqrt{ \frac{1 - t}{1 + t } } = y \\\\\sf then,we~have~to~ find~the~derivative~ of~y~ \\\\\sf Let~ \bf{t = cos \theta},~ then~ \theta = cos^{-1}t\\\\\sf y= {tan}^{ - 1} \sqrt{ \frac{1 - cos \theta}{1 + cos \theta} } \\\\\sf y= {tan}^{ - 1} \sqrt{ \frac{2sin^2 \frac{\theta}{2}}{2 cos^2 \frac{\theta} {2}} } \\\\\sf y= {tan}^{ - 1} \sqrt{ \frac{\cancel{2} sin^2 \frac{\theta}{2}}{\cancel {2} cos^2 \frac{\theta} {2}} }\\\\\sf y= {tan}^{ - 1} \sqrt{ \frac{sin^2 \frac{\theta}{2}}{ cos^2 \frac{\theta} {2}} } \\\\\sf y={tan}^{ - 1} \times \sqrt{tan^2 \frac{\theta }{2} } \\\\\sf y= tan^{-1} \times tan \frac{\theta} {2} \\\\\sf y= \frac{\theta} {2}\\\\\sf y= \frac{cos^{-1}t} {2}\\\\\sf Now, ~derivative ~of ~\bf{y} ~w.r.t ~\bf {t} \\\\\sf \frac{dy} {dt} = \frac{1}{2} \frac{d(cos^{-1}t)} {dt} \\\\\sf = \frac{1}{2} ( \frac{-1}{\sqrt{1-t^2}})$