logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

Differentiate w.r.t. x the function $cot^{-1} \begin{bmatrix} \frac{{\sqrt {1+sin \: x}} + {\sqrt{1-sin \: x}}} {{\sqrt {1+sin \: x}} - {\sqrt{1-sin \: x}}} \\[0.3em] \end{bmatrix}$, where $0 < x < \frac{\pi}{2}$

$\begin{array}{1 1} \frac{3x}{2} \\ \frac{x}{2} \\ \frac{1}{2} \\ \frac{-x}{2} \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $1+\sin x=\cos^2\large\frac{x}{2}$$+\sin^2\large\frac{x}{2}$$+2\cos\large\frac{x}{2}$$\sin\large\frac{x}{2}$
  • $\Rightarrow\big(\cos\large\frac{x}{2}$$+\sin\large\frac{x}{2}\big)^2$
  • $1-\sin x=\big(\cos\large\frac{x}{2}$$-\sin\large\frac{x}{2}\big)^2$
  • $\sqrt{1-\sin x}=\cos\large\frac{x}{2}$$-\sin\large\frac{x}{2}$
  • $\sqrt{1+\sin x}=\cos\large\frac{x}{2}$$+\sin\large\frac{x}{2}$
$y=\cot^{-1}\begin{bmatrix}\Large\frac{(\cos\Large\frac{x}{2}\normalsize+\large\sin\Large\frac{x}{2})+(\cos\Large\frac{x}{2}\normalsize-\large\sin\Large\frac{x}{2})}{(\cos\Large\frac{x}{2}+\sin\Large\frac{x}{2})-(\cos\Large\frac{x}{2}\normalsize-\large\sin\Large\frac{x}{2})}\end{bmatrix}$
$\;\;=\cot^{-1}\begin{bmatrix}\large\frac{2\cos\Large\frac{x}{2}}{2\sin\Large\frac{x}{2}}\end{bmatrix}$
$\;\;=\cot^{-1}(\cot\large\frac{x}{2})$
$\;\;=\large\frac{x}{2}$
answered May 14, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...