Question

cot⁻¹(√1-sinx+√1+sinx/√1-sinx-√1+sinx)=...........(0 < x < π/2)Select Proper option from the given options.
(a) x/2
(b) π/2 - 2x
(c) 2π-x
(d) π- x/2

Answer 1

Dear Student,

Answer: Option D ( π - x/2)

Solution:

cot⁻¹(√1-sinx+√1+sinx/√1-sinx-√1+sinx)

To solve this

First rationalise the denominator

${cot}^{ - 1} (( \frac{ \sqrt{1 - \sin(x) } + \sqrt{1 + \sin(x) } }{ \sqrt{1 - \sin(x) } - \sqrt{1 + \sin(x) } } ) \times ( \frac{ \sqrt{1 - \sin(x) } + \sqrt{1 + \sin(x) } }{ \sqrt{1 - \sin(x) } + \sqrt{1 + \sin(x) } } )) \\ \\ = {cot}^{ - 1} \frac{ ({ \sqrt{1 - \sin(x) } + \sqrt{1 + \sin(x) } })^{2} }{1 - \sin(x) - 1 - \sin(x) } \\ \\ = {cot}^{ - 1} (\frac{1 - \sin(x) + 1 + \sin(x) + 2 \sqrt{1 - {sin}^{2} x} }{ - 2 \sin(x) } )\\ \\ = { \cot}^{ - 1} (\frac{2 + 2 \cos(x) }{ - 2 \sin(x) } ) \\ = {cot}^{ - 1} (\frac{1 + \cos(x) }{ - \sin(x) } ) \ \\ = {cot}^{ - 1} ( \frac{2 {cos}^{2} \frac{x}{2} }{ - 2 \sin( \frac{x}{2} ) \cos( \frac{x}{2} ) } ) \\ \\ = {cot}^{ - 1} ( - \cot( \frac{x}{2} ) \\ \\ = \pi - {cot}^{ - 1} ( \cot( \frac{x}{2} ) \\ \\ = \pi - \frac{x}{2}$

Hope it helps you

Answer 2

${cot}^{ - 1} ( \frac{ \sqrt{1 + sinx} + \sqrt{1 - sinx} }{ \sqrt{1 + sinx} - \sqrt{ 1 - sinx} } ) \\ \\ = {cot}^{ - 1} ( \frac{ \sqrt{( {cos \frac{x}{2} + sin \frac{x}{2} })^{2} } + \sqrt{( {cos \frac{x}{2} - sin \frac{x}{2} })^{2} } }{ \sqrt{( {cos \frac{x}{2} + sin \frac{x}{2} })^{2} } - \sqrt{( {cos \frac{x}{2} - sin \frac{x}{2} })^{2} } } ) \\ \\ = {cot}^{ - 1} ( \frac{cos \frac{x}{2} + sin \frac{x}{2} + cos \frac{x}{2} - sin \frac{x}{2} }{cos \frac{x}{2} + sin \frac{x}{2} - cos \frac{x}{2} + sin \frac{x}{2} } ) \\ \\ = {cot}^{ - 1} ( \frac{2cos \frac{x}{2} }{2sin \frac{x}{2} } ) \\ \\ = {cot}^{ - 1} cot \frac{x}{2} \\ \\ = \frac{x}{2}$