Question

Question 10: Prove cot¯¹ (√1+sinx + √1-sinx /√1+sinx - √1-sinx) = x/2, x ∈(0,π/4 )

Class 12 - Math - Inverse Trigonometric Functions

Answer 1

Hey there!!!


⇒ cot¯¹ (√1+sinx + √1-sinx /√1+sinx - √1-sinx) = x/2   , x ∈(0,π/4 )

⇒ By rationalizing denominator,

⇒ cot¯¹ [ (√1+sinx + √1- sinx)² / (√1+sinx)² - (√1- sinx)²]
⇒ cot¯¹ [ ( 1+ sinx +1 - sinx + 2√1-sin²x ) / (1 + sinx - 1 + sinx) ]
⇒ cot¯¹ [(2 + 2cosx) / 2 sinx]
⇒ cot¯¹ [ 2(1+cosx) / 2sinx]
⇒ cot¯¹ [ (1+cosx) / sinx]

using identity, 1+cos = 2cos²x/2 and sinx = 2 sinx/2.cosx/2

⇒ cot¯¹ [2cos²x/2 / 2 sinx/2.cosx/2]
⇒ cot¯¹( cot x/2)
⇒ x/2

Hence proved!!!

Answer 2

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