Question

The greatest and the least value of ${\left( {{{\sin }^{ - 1}}x} \right)^2} + {\left( {{{\cos }^{ - 1}}x} \right)^2}$ are respectively
$\left( A \right)\,\,\dfrac{{{\pi ^2}}}{4}\,\,and\,\,0$
$\left( B \right)\,\,\dfrac{{{\pi ^2}}}{4}\,\,and\,\,\dfrac{{ - \pi }}{2}$
$\left( C \right)\,\,\dfrac{{5{\pi ^2}}}{4}\,\,and\,\,\dfrac{{{\pi ^2}}}{8}$
$\left( D \right)\,\,\dfrac{{{\pi ^4}}}{4}\,\,and\,\, - \dfrac{{{\pi ^2}}}{4}$

Answer 1

There are some properties or we can say formulas of inverse trigonometry which we can use here. Also, there is one formula of algebra which is used here. But you have to use it carefully as it is applied in a tricky way. We should also know how to find the greatest and the least term when there are some variables involved.

Formula used: ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}$
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$

Complete step by step answer:
In the given question, we have
$\Rightarrow {\left( {{{\sin }^{ - 1}}x} \right)^2} + {\left( {{{\cos }^{ - 1}}x} \right)^2}$
Now, we know that ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
So, applying this formula in above equation
$\Rightarrow {\left( {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \right)^2} - 2{\sin ^{ - 1}}x{\cos ^{ - 1}}x$
As we know that, ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}$
$\Rightarrow {\left( {\dfrac{\pi }{2}} \right)^2} - 2{\sin ^{ - 1}}x\left( {\dfrac{\pi }{2} - {{\sin }^{ - 1}}x} \right)$
On multiplying, we get
$\Rightarrow \dfrac{{{\pi ^2}}}{4} - \pi {\sin ^{ - 1}}x + 2{\left( {{{\sin }^{ - 1}}x} \right)^2}$
Now, taking out $2$ common
$\Rightarrow 2\left[ {{{\left( {{{\sin }^{ - 1}}x} \right)}^2} - \dfrac{\pi }{2}{{\sin }^{ - 1}}x + \dfrac{{{\pi ^2}}}{8}} \right]$
Now adding and subtracting ${\left( {\dfrac{\pi }{4}} \right)^2}$
$\Rightarrow 2\left[ {{{\left( {{{\sin }^{ - 1}}x} \right)}^2} - \dfrac{\pi }{2}{{\sin }^{ - 1}}x + \dfrac{{{\pi ^2}}}{8} + {{\left( {\dfrac{\pi }{4}} \right)}^2} - {{\left( {\dfrac{\pi }{4}} \right)}^2}} \right]$
\Rightarrow 2\left[ {\left\\{ {{{\left( {{{\sin }^{ - 1}}x} \right)}^2} + {{\left( {\dfrac{\pi }{4}} \right)}^2} - \dfrac{\pi }{2}{{\sin }^{ - 1}}x} \right\\} + \left\\{ {\dfrac{{{\pi ^2}}}{8} - {{\left( {\dfrac{\pi }{4}} \right)}^2}} \right\\}} \right]
Now, using ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
$\Rightarrow 2\left[ {{{\left( {{{\sin }^{ - 1}}x - \dfrac{\pi }{4}} \right)}^2} + \dfrac{{{\pi ^2}}}{{16}}} \right]$
We know that ${\left( {{{\sin }^{ - 1}}x - \dfrac{\pi }{4}} \right)^2}$ either can be positive or zero but not negative. So, its minimum value is $0$.
Therefore, the least value of our expression is $2\left( {\dfrac{{{\pi ^2}}}{{16}}} \right)\,i.e.\,\dfrac{{{\pi ^2}}}{8}$ .
To find the greatest value, we have to make ${\left( {{{\sin }^{ - 1}}x - \dfrac{\pi }{4}} \right)^2}$ maximum. To make it maximum we have to add the largest negative value in it which should also lie in the range of ${\sin ^{ - 1}}x$ i.e. $\dfrac{{ - \pi }}{2}$.
Therefore, on putting the value of ${\sin ^{ - 1}}x$
$\Rightarrow 2\left[ {{{\left( { - \dfrac{\pi }{2} - \dfrac{\pi }{4}} \right)}^2} + \dfrac{{{\pi ^2}}}{{16}}} \right]$
$\Rightarrow 2\left[ {\dfrac{{9{\pi ^2}}}{{16}} + \dfrac{{{\pi ^2}}}{{16}}} \right]$
$\Rightarrow \dfrac{{5{\pi ^2}}}{4}$
Therefore, the greatest value of our expression is $\dfrac{{5{\pi ^2}}}{4}$.
Hence, the correct option is $\left( C \right)$.

Note:
The inverse trigonometric functions are also popular as the anti-trigonometric functions. Sometimes these are also termed as arcus functions or cyclometric functions. The inverse trigonometric functions of various trigonometric ratios such as sine, cosine, tangent, cosecant, secant, and cotangent are defined. The inverse trigonometric formulae will help the students to solve the problems in an easy way with the application of these properties.