MCQ
${d \over {dx}}\log \tan \left( {{\pi \over 4} + {x \over 2}} \right) = $
- A$\cos {\rm{ec}}\,x$
- B$ - \cos {\rm{ec}}\,x$
- ✓$\sec x$
- D$ - \sec x$
✓
Answer
Correct option: C.
$\sec x$
c
(c) $\frac{d}{{dx}}\log \tan \left( {\frac{\pi }{4} + \frac{x}{2}} \right) = \frac{1}{{\tan \left( {\frac{\pi }{4} + \frac{x}{2}} \right)}}{\sec ^2}\left( {\frac{\pi }{4} + \frac{x}{2}} \right).\frac{1}{2}$
(c) $\frac{d}{{dx}}\log \tan \left( {\frac{\pi }{4} + \frac{x}{2}} \right) = \frac{1}{{\tan \left( {\frac{\pi }{4} + \frac{x}{2}} \right)}}{\sec ^2}\left( {\frac{\pi }{4} + \frac{x}{2}} \right).\frac{1}{2}$
$ = \frac{1}{2}.\frac{1}{{\sin \left( {\frac{\pi }{4} + \frac{x}{2}} \right)\cos \left( {\frac{\pi }{4} + \frac{x}{2}} \right)}} = \frac{1}{{\sin \left( {\frac{\pi }{2} + x} \right)}} = \frac{1}{{\cos x}} = \sec x$.
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