The value of -sin x-1- sin x-2-sin x-3- -sin x- in - 3-2- is

The correct option is C 2 1 ≤sin x ≤ 0 when x ∈(π, 3π2] Hence, [1+sin x] = 0 and [sin x] = 1 ∴ [sin x] + [1+sin x] + 2+ [sin ...

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First Fundamental Theorem of Calculus

The value of ...

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The value of $[sin x] + [1 + sin x] + 2 + [sin x] + 3 + [sin x]$ in $(π, \frac{3 π}{2}]$ is

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

(0 < x < \frac{π}{2})

\frac{π}{2} < x < \frac{3 π}{2}

\sqrt{\frac{1 - sin x}{1 + sin x}}

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

(0 < x < \frac{π}{2})

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

\frac{d y}{d x}

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

S h o w t h a t

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}

f o r x \in (0, \frac{π}{2})

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