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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
Differentiate w.r.t. x the function in Exercise:
$\cot^{-1}\Big[\frac{\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\text{x}}}\Big],\ 0<\text{x}<\frac{\pi}{2}$
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Answer

Let $\text{y}=\cot^{-1}\Big[\frac{\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\text{x}}}\Big]\ \dots(1)$
Then, $\frac{\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}}}{\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\text{x}}}$
$=\frac{(\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}})^2}{(\sqrt{1+\sin\text{x}}-\sqrt{1-\sin\text{x}})(\sqrt{1+\sin\text{x}}+\sqrt{1-\sin\text{x}})}$
$=\frac{(1+\sin\text{x})+(1-\sin\text{x})+2\sqrt{(1-\sin\text{x})(1+\sin\text{x})}}{(1+\sin\text{x})-(1-\sin\text{x})}$
$=\frac{2+2\sqrt{1-\sin^2\text{x}}}{2\sin\text{x}}$
$=\frac{1+\cos\text{x}}{\sin\text{x}}$
$=\frac{2\cos^2\frac{\text{x}}{2}}{2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}$
$=\cot\frac{\text{x}}{2}$
Therefore, equation (1) becomes
$\text{y}=\cot^{-1}\Big(\cot\frac{\text{x}}{2}\Big)$
$\Rightarrow\ \text{y}=\frac{\text{x}}{2}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{x)}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{2}$

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