Question

# $$\displaystyle\frac{dy}{dx}$$ for $$y=\tan^{-1}\left\{\sqrt{\displaystyle\frac{1+\cos x}{1-\cos x}}\right\}$$, where $$0 < x < \pi$$, is?

A
12
B
0
C
1
D
1

Solution

## The correct option is C $$\dfrac{-1}{2}$$$$To\; find\; \cfrac { d }{ dx } \left( \tan ^{ -1 }{ \sqrt { \cfrac { 1+\cos { x } }{ 1-\cos { x } } } } \right) \\ By\; Applying\; chain\; rule\; let\; us\; take\; u=\sqrt { \cfrac { 1+\cos { x } }{ 1-\cos { x } } } \\ Therefore\; \cfrac { d }{ du } \left( \tan ^{ -1 }{ u } \right) \cdot \cfrac { d }{ dx } \left( \sqrt { \cfrac { 1+\cos { x } }{ 1-\cos { x } } } \right) \\ Step\; 1:\; \cfrac { d }{ du } \tan ^{ -1 }{ u } =\cfrac { 1 }{ { u }^{ 2 }+1 } \\ Step\; 2:\; \cfrac { d }{ dx } \sqrt { \cfrac { 1+\cos { x } }{ 1-\cos { x } } } =-\cfrac { \sin { x } }{ \sqrt { 1+\cos { x } } \cdot ({ 1-\cos { x) } }^{ 3/2 } } \\ Combining\; Step\; 1\; and\; Step\; 2\\ =\cfrac { 1 }{ { u }^{ 2 }+1 } \cdot (-\cfrac { \sin { x } }{ \sqrt { 1+\cos { x } } \cdot ({ 1-\cos { x) } }^{ 3/2 } } )\\ substituting\; the\; value\; of\; 'u'\; in\; the\; given\; equation\\ =\cfrac { 1 }{ { (\sqrt { \cfrac { 1+\cos { x } }{ 1-\cos { x } } } ) }^{ 2 }+1 } \cdot (-\cfrac { \sin { x } }{ \sqrt { 1+\cos { x } } \cdot ({ 1-\cos { x) } }^{ 3/2 } } )\\ =-\cfrac { sinx }{ 2\sqrt { 1+cosx } (1-{ cosx) }^{1/2 } } \\ Since\; 0<\; x\; <\pi \\ Let\; us\; take\; the\; value\; as\; \cfrac { \pi }{ 2 } \\ So\; the\; value\; would\; be\; \\ =-\cfrac { \sin { \cfrac { \pi }{ 2 } } }{ 2\; \sqrt { 1+\cos { \cfrac { \pi }{ 2 } } } \; { \left( 1-\cos { \cfrac { \pi }{ 2 } } \right) }^{ \cfrac { 1 }{ 2 } } } \\ =-\cfrac { 1 }{ 2\sqrt { 1+0 } \; ({ 1-0 })^{1/2 }\; } \\ =-\cfrac { 1 }{ 2 } \\ Therefore\; \cfrac { d }{ dx } \left( \tan ^{ -1 }{ \sqrt { \cfrac { 1+\cos { x } }{ 1-\cos { x } } } } \right) =-\cfrac { 1 }{ 2 } \; at\; 0<\; x\; <\pi$$Hence, $$\dfrac{-1}{2}$$ is the correct answer.Mathematics

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