Inverse Trigonometric Functions — Class 12 Maths Solution

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 $x = 0, - 1$