Question

If $y = {\sec ^{ - 1}}\left[ {\text{cosec }x} \right] + \text{cosec }^{ - 1}\left[ {\sec x} \right] + {\sin ^{ - 1}}\left[ {\cos x} \right] + {\cos ^{ - 1}}\left[ {\sin x} \right]$, then calculate $\dfrac{{dy}}{{dx}}$.
A. 0
B. 2
C. -2
D. -4

Answer 1

Hint: First apply complementary angles in trigonometry ratios to simplify the given equation. Then apply the derivative formula to the equation to find $\dfrac{{dy}}{{dx}}$.

Formula Used:
Trigonometry ratios of the complementary angle
$\sin \theta = \cos \left( {{{90}^ \circ } - \theta } \right)$
$\cos \theta = \sin \left( {{{90}^ \circ } - \theta } \right)$
$\sec \theta = \text{cosec }\left( {{{90}^ \circ } - \theta } \right)$
$\text{cosec }\theta = \sec \left( {{{90}^ \circ } - \theta } \right)$
${\sec ^{ - 1}}\left( {\sec \theta } \right) = \theta $
$\text{cosec }^{ - 1}\left( {\text{cosec }\theta } \right) = \theta $
${\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta $
${\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta $
$\dfrac{d}{{dx}}\left( {f\left( x \right) \pm g\left( x \right)} \right) = \dfrac{d}{{dx}}f\left( x \right) \pm \dfrac{d}{{dx}}g\left( x \right)$
$\dfrac{d}{{dx}}\left( {mx} \right) = m$ where $m$ is constant.
$\dfrac{d}{{dx}}\left( c \right) = 0$ where $c$ is constant

Complete step by step solution:
Given equation is
$y = {\sec ^{ - 1}}\left[ {\text{cosec }x} \right] + \text{cosec }^{ - 1}\left[ {\sec x} \right] + {\sin ^{ - 1}}\left[ {\cos x} \right] + {\cos ^{ - 1}}\left[ {\sin x} \right]$
we will apply trigonometry ratios of complementary angles $\sin \theta = \cos \left( {{{90}^ \circ } - \theta } \right)$, $\cos \theta = \sin \left( {{{90}^ \circ } - \theta } \right)$, $\sec \theta = \cos ec\left( {{{90}^ \circ } - \theta } \right)$ and $\text{cosec }\theta = \sec \left( {{{90}^ \circ } - \theta } \right)$.
$y = {\sec ^{ - 1}}\left[ {\sec \left( {{{90}^ \circ } - x} \right)} \right] + \text{cosec }^{ - 1}\left[ {\text{cosec }\left( {{{90}^ \circ } - x} \right)} \right] + {\sin ^{ - 1}}\left[ {\sin \left( {{{90}^ \circ } - x} \right)} \right] + {\cos ^{ - 1}}\left[ {\cos \left( {{{90}^ \circ } - x} \right)} \right]$
Applying the formulas ${\sec ^{ - 1}}\left( {\sec \theta } \right) = \theta $ , $\text{cosec }^{ - 1}\left( {\text{cosec }\theta } \right) = \theta $, ${\sin ^{ - 1}}\left( {\sin \theta } \right) = \theta $, and ${\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta $
$ \Rightarrow y = \left( {{{90}^ \circ } - x} \right) + \left( {{{90}^ \circ } - x} \right) + \left( {{{90}^ \circ } - x} \right) + \left( {{{90}^ \circ } - x} \right)$
$ \Rightarrow y = {360^ \circ } - 4x$
We know that, ${360^ \circ } = 2\pi $.
$ \Rightarrow y = 2\pi - 4x$
Differentiate with respect to $x$
 $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {2\pi - 4x} \right)$
Apply the formula $\dfrac{d}{{dx}}\left( {f\left( x \right) \pm g\left( x \right)} \right) = \dfrac{d}{{dx}}f\left( x \right) \pm \dfrac{d}{{dx}}g\left( x \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {2\pi } \right) - \dfrac{d}{{dx}}\left( {4x} \right)$
Now applying the formula $\dfrac{d}{{dx}}\left( {mx} \right) = m$ and $\dfrac{d}{{dx}}\left( c \right) = 0$
$ \Rightarrow \dfrac{{dy}}{{dx}} = 0 - 4$
$ \Rightarrow \dfrac{{dy}}{{dx}} = - 4$

Option ‘D’ is correct

Note: Students often start to solve problems directly. They did not apply the complimentary angles of trigonometry ratios to simplify $y = {\sec ^{ - 1}}\left[ {\text{cosec }x} \right] + \text{cosec }^{ - 1}\left[ {\sec x} \right] + {\sin ^{ - 1}}\left[ {\cos x} \right] + {\cos ^{ - 1}}\left[ {\sin x} \right]$. It becomes lengthy and complicated. So, first we will apply the complementary angles of trigonometry ratios and formulas inverse function to simplify the equation. Then we derive it.