-1-1-sin x-1-sin x-1-sin x x-1-sin x-x-2- x -0- -4

Given, cot^ 1 ( √(1+sin x)+√(1 sin x)/√(1+sin x) √(1 sin x) )=x/2, xϵ ( 0, π/4 ) LHS=Cot^ 1 ( √(1+sin x)+√(1 sin x)/√(1+sin x) √(1 sin x) )=x/2, xϵ ( 0, π/4 ) ...

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Byju's Answer

Standard XII

Mathematics

Basic Inverse Trigonometric Functions

-1√1+sin x+√1...

Question

$c o t^{- 1} (\frac{\sqrt{1 + s i n x} + \sqrt{1 - s i n x}}{\sqrt{1 + s i n x} - \sqrt{1 - s i n x}}) = \frac{x}{2}, x ϵ (0, \frac{π}{4})$

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Solution

Given, $c o t^{- 1} (\frac{\sqrt{1 + s i n x} + \sqrt{1 - s i n x}}{\sqrt{1 + s i n x} - \sqrt{1 - s i n x}}) = \frac{x}{2}, x ϵ (0, \frac{π}{4})$

$L H S = C o t^{- 1} (\frac{\sqrt{1 + s i n x} + \sqrt{1 - s i n x}}{\sqrt{1 + s i n x} - \sqrt{1 - s i n x}}) = \frac{x}{2}, x ϵ (0, \frac{π}{4})$

Now, we can write
$\sqrt{1 + s i n x} = \sqrt{s i n^{2} \frac{x}{2} + c o s^{2} \frac{x}{2} + 2 s i n \frac{x}{2} c o s \frac{x}{2}} = \sqrt{{(s i n \frac{x}{2} + c o s \frac{x}{2})}^{2}} [∵ s i n^{2} \frac{x}{2} + c o s^{2} \frac{x}{2} = 1 a n d s i n x = 2 s i n \frac{x}{2} c o s \frac{x}{2}] \Rightarrow \sqrt{1 + s i n x} = s i n \frac{x}{2} + c o s \frac{x}{2}$

similarly, we can get $\sqrt{1 - s i n x} = c o s \frac{x}{2} - s i n \frac{x}{2}$
On substituting above two values in Eq.(i), we get
$L H S = c o t^{- 1} (\frac{s i n \frac{x}{2} + c o s \frac{x}{2} + c o s \frac{x}{2} - s i n \frac{x}{2}}{s i n \frac{x}{2} + c o s \frac{x}{2} + s i n \frac{x}{2} - c o s \frac{x}{2}}) = c o t^{- 1} (\frac{2 c o s \frac{x}{2}}{2 s i n \frac{x}{2}}) = c o t^{- 1} (c o t \frac{x}{2}) = \frac{x}{2} = R H S$

Alternative method
$L H S = c o t^{- 1} [\frac{\sqrt{1 + s i n x} + \sqrt{1 - s i n x}}{\sqrt{1 + s i n x} - \sqrt{1 - s i n x}}] = c o t^{- 1} [\frac{\sqrt{1 + s i n x} + \sqrt{1 - s i n x}}{\sqrt{1 + s i n x} - \sqrt{1 - s i n x}} \times \frac{\sqrt{1 + s i n x} + \sqrt{1 - s i n x}}{\sqrt{1 + s i n x} + \sqrt{1 - s i n x}}]$
(by rationalyzing denominator)
$= c o t^{- 1} [\frac{(\sqrt{1 + s i n x} + \sqrt{1 - s i n x})^{2}}{(\sqrt{1 + s i n x})^{2} - (\sqrt{1 - s i n x})^{2}}] = c o t^{- 1} [\frac{1 + s i n x + 1 - s i n x + 2 \sqrt{1 - s i n^{2} x}}{1 + s i n x - 1 + s i n x}] = c o t^{- 1} [\frac{2 + 2 c o s x}{2 s i n x}] [∵ c o s^{2} x + s i n^{2} x = 1 \Rightarrow c o s x = \sqrt{1 - s i n^{2} x}] = c o t^{- 1} [\frac{1 + c o s x}{s i n x}] = c o t^{- 1} [\frac{2 c o s^{2} \frac{x}{2}}{2 s i n \frac{x}{2} c o s \frac{x}{2}}] U s i n g t h e s t a n d a r d f o r m u l a e w e g e t, = c o t^{- 1} [c o t \frac{x}{2}] = \frac{x}{2} = R H S$

Note: If $x ϵ (0, \frac{π}{2})$ , then $\sqrt{1 - s i n x} = c o s \frac{x}{2} - s i n \frac{x}{2}$ and if $x ε (\frac{π}{2}, π)$ , then $\sqrt{1 - s i n x} = s i n \frac{x}{2} - c o s \frac{x}{2}$

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Similar questions

Q. Prove:

{cot}^{- 1} (\frac{\sqrt{1 + sin x} + \sqrt{1 - sin x}}{\sqrt{1 + sin x} - \sqrt{1 - sin x}}) = \frac{x}{2}, x \in (0, \frac{π}{4})

{cot}^{- 1} (\frac{\sqrt{1 - sin x} + \sqrt{1 + sin x}}{\sqrt{1 - sin x} - \sqrt{1 + sin x}})

=....

(0 < x < \frac{π}{2})

Q. If

{cot}^{- 1} (\frac{\sqrt{1 + sin x} + \sqrt{1 - sin x}}{\sqrt{1 + sin x} - \sqrt{1 - sin x}}) = \frac{x}{m}, x \in (0, \frac{π}{4})

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Q. Simplify

{cot}^{- 1} [\frac{\sqrt{1 + sin x} + \sqrt{1 - sin x}}{\sqrt{1 + sin x} - \sqrt{1 - sin x}}] = \frac{x}{m}; x \in (0, \frac{π}{4})

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Q. If

y = {cot}^{- 1} \frac{\sqrt{1 + sin x} + \sqrt{1 - sin x}}{\sqrt{1 + sin x} - \sqrt{1 - sin x}}

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x \in (0, \frac{π}{2}) \cup (\frac{π}{2}, π)

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cot−1(√1+sin x+√1−sin x√1+sin x−√1−sin x)=x2,xϵ(0,π4)

$c o t^{- 1} (\frac{\sqrt{1 + s i n x} + \sqrt{1 - s i n x}}{\sqrt{1 + s i n x} - \sqrt{1 - s i n x}}) = \frac{x}{2}, x ϵ (0, \frac{π}{4})$