Question

If we have an expression ${\sin ^{ - 1}}\left( {\dfrac{x}{{13}}} \right) + \cos e{c^{ - 1}}\left( {\dfrac{{13}}{{12}}} \right) = \dfrac{\pi }{2}$, then the value of x is:
(A) $5$
(B) $4$
(C) $12$
(D) $11$

Answer 1

In the given problem, we are required to find the value of x from the given equation involving the inverse trigonometric functions. So, to solve the equation, we firstly take sin on both sides of the equation and calculate sine and cosine functions of an angle whose cosecant is given to us. Such problems require basic knowledge of trigonometric ratios and formulae. Besides this, knowledge of concepts of inverse trigonometry is extremely essential to answer these questions correctly.

Complete step-by-step solution:
So, In the given problem, we have to find the value of x in ${\sin ^{ - 1}}\left( {\dfrac{x}{{13}}} \right) + \cos e{c^{ - 1}}\left( {\dfrac{{13}}{{12}}} \right) = \dfrac{\pi }{2}$.
Taking $\cos e{c^{ - 1}}\left( {\dfrac{{13}}{{12}}} \right)$ to the right side of the equation, we get,
$\Rightarrow {\sin ^{ - 1}}\left( {\dfrac{x}{{13}}} \right) = \dfrac{\pi }{2} - \cos e{c^{ - 1}}\left( {\dfrac{{13}}{{12}}} \right)$
Now, we know that secant and cosecant functions are complimentary. So, $\left( {\dfrac{\pi }{2} - \cos e{c^{ - 1}}x} \right) = {\sec ^{ - 1}}x$. Hence, we get,
$\Rightarrow {\sin ^{ - 1}}\left( {\dfrac{x}{{13}}} \right) = {\sec ^{ - 1}}\left( {\dfrac{{13}}{{12}}} \right)$
So, taking sine on both sides of the equation, we get,
$\Rightarrow \sin \left( {{{\sin }^{ - 1}}\left( {\dfrac{x}{{13}}} \right)} \right) = \sin \left( {{{\sec }^{ - 1}}\left( {\dfrac{{13}}{{12}}} \right)} \right)$
Now, we know that $\sin \left( {{{\sin }^{ - 1}}\left( x \right)} \right) = x$. So, we get,
$\Rightarrow \left( {\dfrac{x}{{13}}} \right) = \sin \left[ {{{\sec }^{ - 1}}\left( {\dfrac{{13}}{{12}}} \right)} \right]$
Now, we have to find sine of the angle whose secant is given to us as $\left( {\dfrac{{13}}{{12}}} \right)$.
Let us assume $\theta$ to be the concerned angle.
Then, $\theta = {\sec ^{ - 1}}\left( {\dfrac{{13}}{{12}}} \right)$
Taking secant on both sides of the equation, we get
$\Rightarrow \sec \theta = \left( {\dfrac{{13}}{{12}}} \right)$
To evaluate the value of the required expression, we must keep in mind the formulae of basic trigonometric ratios.
We know that, $\sec \theta = \dfrac{\text{Hypotenuse}}{\text{Base}}$and $\sin \theta = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$.
So, $\sec \theta = \dfrac{\text{Hypotenuse}}{\text{Base}} = \dfrac{{13}}{{12}}$
Let the length of the hypotenuse be $13x$.
Then, length of base $= 12x$.
Now, applying Pythagoras Theorem,
${\left( \text{Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( \text{Perpendicular} \right)^2}$
$= {\left( {13x} \right)^2} = {\left( {12x} \right)^2} + {\left( \text{Perpendicular} \right)^2}$
$= 169{x^2} = 144{x^2} + {\left( \text{Perpendicular} \right)^2}$
$= {\left( \text{Perpendicular} \right)^2} = \left( {169 - 144} \right){x^2}$
$= \left( \text{Perpendicular} \right) = \sqrt {25{x^2}}$
$= \left( \text{Perpendicular} \right) = 5x$
So, we get $\left( \text{Perpendicular}\right) = 5x$
Hence, $\sin \theta = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{{5x}}{{13x}}$
$\Rightarrow \sin \theta = \dfrac{5}{{13}}$
So, the value of$\sin \left[ {{{\sec }^{ - 1}}\left( {\dfrac{{13}}{{12}}} \right)} \right]$ is $\left( {\dfrac{5}{{13}}} \right)$.
Therefore, we get the equation as,
$\Rightarrow \left( {\dfrac{x}{{13}}} \right) = \left( {\dfrac{5}{{13}}} \right)$
Cross multiplying the terms, we get,
$\Rightarrow x = \left( {\dfrac{5}{{13}}} \right) \times 13$
Cancelling the common factors in numerator and denominator, we get,
$\Rightarrow x = 5$
So, the value of x is $5$. Hence, option (A) is the correct answer.

Note: For finding a trigonometric ratio for an angle given in terms of an inverse trigonometric ratio, we have to first assume that angle to be some unknown, let's say $\theta$. Then proceeding further, we have to find the value of a trigonometric function of that unknown angle $\theta$. Then we find the required trigonometric ratio with help of basic trigonometric formulae and definitions of trigonometric ratios. Such questions require clarity of basic concepts of trigonometric functions as well as their inverse. Care should be taken while doing calculations.