Question

${lim}_{x\to \frac{\pi }{2}}\frac{(1-\tan \frac{x}{2})(1-\sin x)}{[1+\tan \frac{x}{2}][\pi -2x{]}^3}$

Answer 1

$\begin{array}{c}{lt}_{x\to \frac{\pi }{2}}\frac{\left(1-\tan \frac{x}{2}\right)\left(1-\sin x\right)}{\left(1+\tan \frac{x}{2}\right){\left(\pi -2x\right)}^3}\\ ={lt}_{x\to \frac{\pi }{2}}\frac{\tan \left(\frac{\pi }{4}-\frac{x}{2}\right)\left(1-\sin x\right)}{{\left(\pi -2x\right)}^3}\\ ={lt}_{x\to \frac{\pi }{2}}\frac{\left(\sin x-1\right)\tan \left(\frac{\pi }{4}-\frac{x}{2}\right)}{{\left(2x-\pi \right)}^3}\\ Letx=\frac{\pi }{2}+h\\ Now,\\ {lt}_{h\to 0}\frac{-\left(\cosh -1\right)\tan \left(\frac{h}{2}\right)}{{\left(2h\right)}^3}\\ ={lt}_{h\to 0}\frac{\left(1-\cosh \right)\tan \left(\frac{h}{2}\right)}{8h^3}\\ =\frac{1}{8}{lt}_{h\to 0}\left(\frac{1-\cosh }{h^2}\right)\frac{\tan \left(\frac{h}{2}\right)}{h}\\ ={lt}_{h\to 0}\frac{\tan \left(\frac{h}{2}\right)}{\frac{h}{2}\times 2}\\ =\frac{1}{32}\end{array}$

which is the required answer.