Find the real Solution of
${\tan ^{ – 1}}\sqrt {x(x + 1)} + {\sin ^{ – 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{2}$.
[NCERT,Exemplar.2.3,Q.7,Page.36]

Find the real Solution of
${\tan ^{ – 1}}\sqrt {x(x + 1)} + {\sin ^{ – 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{2}$.
[NCERT,Exemplar.2.3,Q.7,Page.36]

We have, ${\tan ^{ – 1}}\sqrt {x(x + 1)} + {\sin ^{ – 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{2}$

……(i)
Let ${\sin ^{ – 1}}\sqrt {{x^2} + x + 1} = \theta $

$ \Rightarrow $$\sin \theta = \sqrt {\frac{{{x^2} + x + 1}}{1}} $

$ \Rightarrow $$\tan \theta = \frac{{\sqrt {{x^2} + x + 1} }}{{\sqrt { – {x^2} – x} }}$

$ = {\sin ^{ – 1}}\sqrt {{x^2} + x + 1} $
On putting the value of $\theta $ in Eq. (i),

we get ${\tan ^{ – 1}}\sqrt {x(x + 1)} + {\tan ^{ – 1}}\frac{{\sqrt {{x^2} + x + 1} }}{{\sqrt { – {x^2} – x} }} = \frac{\pi }{2}$

We know that, ${\tan ^{ – 1}}x + {\tan ^{ – 1}}y = {\tan ^{ – 1}}\left( {\frac{{x + y}}{{1 – xy}}} \right),xy < 1$

$therefore, {\tan ^{ – 1}}\left[ {\frac{{\sqrt {x(x + 1)} + \sqrt {\frac{{{x^2} + x + 1}}{{ – {x^2} – x}}} }}{{1 – \sqrt {x(x + 1)} \cdot \sqrt {\frac{{{x^2} + x + 1}}{{ – {x^2} – x}}} }}} \right] = \frac{\pi }{2}$

$ \Rightarrow $${\tan ^{ – 1}}\left[ {\frac{{\sqrt {{x^2} + x} + \sqrt {\frac{{{x^2} + x + 1}}{{ – 1\left( {{x^2} + x} \right)}}} }}{{1 – \sqrt {\left( {{x^2} + x} \right) \cdot \frac{{\left( {{x^2} + x + 1} \right)}}{{ – 1\left( {{x^2} + x} \right)}}} }}} \right] = \frac{\pi }{2}$

$ \Rightarrow $$\frac{{{x^2} + x + \sqrt { – \left( {{x^2} + x + 1} \right)} }}{{\left[ {1 – \sqrt { – \left( {{x^2} + x + 1} \right.} } \right]\sqrt {\left( {{x^2} + x} \right)} }} = \tan \frac{\pi }{2} = \frac{1}{0}$

$ \Rightarrow \left[ {1 – \sqrt { – \left( {{x^2} + x + 1} \right)} } \right]\sqrt {\left( {{x^2} + x} \right)} = 0$

$ \Rightarrow – \left( {{x^2} + x + 1} \right) = 1$ or ${x^2} + x = 0$
$ \Rightarrow – {x^2} – x – 1 = 1$ or $x(x + 1) = 0$

$ \Rightarrow $${x^2} + x + 2 = 0$ or $x(x + 1) = 0$
$therefore, x = \frac{{ – 1 \pm \sqrt {1 – 4 \times 2} }}{2}$

$ \Rightarrow $$x = 0$ or $x = – 1$
For real Solution, we have $x = 0, – 1$

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