If y - x-a2 - 1 - 2-a2 - 1tan-1 sin xa- -a2 - 1 - cos x where a- - -1 - 1- - then y- -2 is equal to

Let √(a^2 1) = C_1 and nbsp;a + √(a^2 1) = C_2 ⇒ C_2 C_1 = a ∴ y = xC_1 2C_1tan^ 1 (sin xC_2 + cos x) y'= 1C_1 2C_1·11+(sin xC_2+cos x)^2·(C_2+cos x) ...

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Byju's Answer

Standard XII

Mathematics

Composite Function

If y = x√a2 -...

Question

If $y = \frac{x}{\sqrt{a^{2} - 1}} - \frac{2}{\sqrt{a^{2} - 1}} {tan}^{- 1} (\frac{sin x}{a + \sqrt{a^{2} - 1} + cos x})$ where $a \in (- \infty, - 1) \cup (1, \infty),$ then $y^{'} (\frac{π}{2})$ is equal to

\frac{1}{a}

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\frac{2}{a}

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\frac{1}{2 a}

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a

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Solution

The correct option is A $\frac{1}{a}$
Let $\sqrt{a^{2} - 1} = C_{1}$ and $a + \sqrt{a^{2} - 1} = C_{2}$
$\Rightarrow C_{2} - C_{1} = a$
$∴ y = \frac{x}{C_{1}} - \frac{2}{C_{1}} {tan}^{- 1} (\frac{sin x}{C_{2} + cos x})$

$y^{'} = \frac{1}{C_{1}} - \frac{2}{C_{1}} \cdot \frac{1}{1 + {(\frac{sin x}{C_{2} + cos x})}^{2}} \cdot \frac{(C_{2} + cos x) cos x - sin x (- sin x)}{(C_{2} + cos x)^{2}} = \frac{1}{C_{1}} - \frac{2}{C_{1}} \cdot \frac{C_{2} cos x + 1}{(C_{2} + cos x)^{2} + {sin}^{2} x} {\frac{d y}{d x} \begin{matrix} \end{matrix} ∣ ∣ ∣}_{x = \frac{π}{2}} = \frac{1}{C_{1}} - \frac{2}{C_{1}} \cdot \frac{1}{(C_{2})^{2} + 1} = \frac{1}{\sqrt{a^{2} - 1}} - \frac{2}{\sqrt{a^{2} - 1}} \cdot \frac{1}{2 a^{2} - 1 + 2 a \sqrt{a^{2} - 1} + 1} = \frac{2 a^{2} + 2 a \sqrt{a^{2} - 1} - 2}{\sqrt{a^{2} - 1} (2 a^{2} + 2 a \sqrt{a^{2} - 1})}$
$= \frac{2 a^{2} + 2 a \sqrt{a^{2} - 1} - 2}{\sqrt{a^{2} - 1} (2 a (a + \sqrt{a^{2} - 1}))} = \frac{2 (a^{2} - 1) + 2 a \sqrt{a^{2} - 1}}{\sqrt{a^{2} - 1} (2 a (a + \sqrt{a^{2} - 1}))} = \frac{2 \sqrt{a^{2} - 1} (\sqrt{a^{2} - 1} + a)}{(\sqrt{a^{2} - 1}) (2 a) (a + \sqrt{a^{2} - 1})} = \frac{1}{a}$

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If y=x√a2−1−2√a2−1tan−1(sinxa+√a2−1+cosx) where a∈(−∞,−1)∪(1,∞), then y′(π2) is equal to

If $y = \frac{x}{\sqrt{a^{2} - 1}} - \frac{2}{\sqrt{a^{2} - 1}} {tan}^{- 1} (\frac{sin x}{a + \sqrt{a^{2} - 1} + cos x})$ where $a \in (- \infty, - 1) \cup (1, \infty),$ then $y^{'} (\frac{π}{2})$ is equal to