class 12 maths continuity and differentiability

${\cot ^{ - 1}}\left[ {\cfrac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right],0 < x < \cfrac{\pi }{2}.$

VAVidaara Admin Asked 9d ago 0 views 0 answers
📘 Continuity and Differentiability NCERT Misc. ,Q.6,Page 191 SA

${\cot ^{ - 1}}\left[ {\cfrac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right],0 < x < \cfrac{\pi }{2}.$

Official Solution

VVidaara Team ✓ Verified solution NCERT & Exemplar

Let $y = {\cot ^{ - 1}}\left[ {\cfrac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right],0 < x < \cfrac{\pi }{2}$

$\Rightarrow$ $y = {\cos ^{ - 1}}\left[ {\cfrac{{\sqrt {{{\cos }^2}\cfrac{x}{2} + {{\sin }^2}\cfrac{x}{2} + 2\sin \cfrac{x}{2}\cos \cfrac{x}{2}} + \sqrt {{{\cos }^2}\cfrac{x}{2} + {{\sin }^2}\cfrac{x}{2} - 2\sin \cfrac{x}{2}\cos \cfrac{x}{2}} }}{{\sqrt {{{\cos }^2}\cfrac{x}{2} + {{\sin }^2}\cfrac{x}{2} + 2\sin \cfrac{x}{2}\cos \cfrac{x}{2}} - \sqrt {{{\cos }^2}\cfrac{x}{2} + {{\sin }^2}\cfrac{x}{2} + 2\sin \cfrac{x}{2}\cos \cfrac{x}{2}} }}} \right]$

$\Rightarrow$ $y = {\cot ^{ - 1}}\left[ {\cfrac{{\sqrt {{{\left( {\cos \cfrac{x}{2} + \sin \cfrac{x}{2}} \right)}^2}} + \sqrt {{{\left( {\cos \cfrac{x}{2} - \sin \cfrac{x}{2}} \right)}^2}} }}{{\sqrt {{{\left( {\cos \cfrac{x}{2} - \sin \cfrac{x}{2}} \right)}^2}} - \sqrt {{{\left( {\cos \cfrac{x}{2} - \sin \cfrac{x}{2}} \right)}^2}} }}} \right]$

$\Rightarrow$ $y = {\cot ^{ - 1}}\left[ {\cfrac{{\cos \cfrac{x}{2} + \sin \cfrac{x}{2} + \cos \cfrac{x}{2} - \sin \cfrac{x}{2}}}{{\cos \cfrac{x}{2} + \sin \cfrac{x}{2} - \cos \cfrac{x}{2} + \sin \cfrac{x}{2}}}} \right]$

$\Rightarrow$ $y = {\cot ^{ - 1}}\left[ {\cfrac{{2\cos \cfrac{x}{2}}}{{2\sin \cfrac{x}{2}}}} \right]$

$\Rightarrow$ $y = {\cot ^{ - 1}}\left[ {\cot \cfrac{x}{2}} \right]y = \cfrac{x}{2} \Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{1}{2}$

View the full step-by-step solution page & related questions →

Community Answers (0)

Log in to post your own answer or join the discussion.

Discussion (0)

No comments yet — start the discussion.

← Back to all questions