Question

Solve ${{\sin }^{-1}}\dfrac{5}{x}+{{\sin }^{-1}}\dfrac{12}{x}=\dfrac{\pi }{2}$

Answer 1

Hint: In the question, we are given an equation that involves inverse trigonometric functions. First, we will learn what is inverse trigonometric functions, how it is represented and some important properties of that. Then we will take one consideration and we will use this in our equation to get our required answer.

Complete answer:
In the question we are given an equation which has inverse trigonometric functions.
Inverse trigonometric functions:
Inverse trigonometric functions are the inverse functions of trigonometric functions
Inverse function of $\sin x$ is represented as ${{\sin }^{-1}}x$
Inverse function of $\cos x$ is represented as ${{\cos }^{-1}}x$
Inverse function of $\tan x$ is represented as ${{\tan }^{-1}}x$
Inverse function of $\text{cosec}x$ is represented as $\text{cose}{{\text{c}}^{-1}}x$
Inverse function of $\sec x$ is represented as ${{\sec }^{-1}}x$
Inverse function of $\tan x$ is represented as ${{\tan }^{-1}}x$
The most important properties of inverse trigonometric functions are
$\sin \left( {{\sin }^{-1}}x \right)=x$ And ${{\sin }^{-1}}\left( \sin x \right)=x$
$\cos \left( {{\cos }^{-1}}x \right)=x$ And ${{\cos }^{-1}}\left( \cos x \right)=x$
$\tan \left( {{\tan }^{-1}}x \right)=x$ And ${{\tan }^{-1}}\left( \tan x \right)=x$
We can generalize this properties to all trigonometric functions.
Now we will proceed to our question
Our given equations is
${{\sin }^{-1}}\dfrac{5}{x}+{{\sin }^{-1}}\dfrac{12}{x}=\dfrac{\pi }{2}$
Let us consider $\alpha ={{\sin }^{-1}}\dfrac{12}{x}$
We can also write it as
$\sin \alpha =\dfrac{12}{x}$
As we know ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
From this trigonometry identity we can say that
${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $
Now we will use this in our equation
$\sin \alpha =\dfrac{12}{x}$
Using trigonometry identity and this value we will find value of $\cos \alpha $
So,
${{\cos }^{2}}\alpha =1-{{\sin }^{2}}\alpha $
Substituting $\sin \alpha =\dfrac{12}{x}$
${{\cos }^{2}}\alpha =1-{{\sin }^{2}}\alpha$
$\Rightarrow {{\cos }^{2}}\alpha =1-{{\left( \dfrac{12}{x} \right)}^{2}}$
Now we will simplify this
$ {{\cos }^{2}}\alpha =1-{{\left( \dfrac{12}{x} \right)}^{2}} $
$ \Rightarrow {{\cos }^{2}}\alpha =1-\dfrac{{{\left( 12 \right)}^{2}}}{{{x}^{2}}}$
$ \Rightarrow {{\cos }^{2}}\alpha =1-\dfrac{144}{{{x}^{2}}}$
Taking LCM on right hand side
$\Rightarrow {{\cos }^{2}}\alpha =\dfrac{{{x}^{2}}-144}{{{x}^{2}}}$
Now we will take square root both sides
$\sqrt{{{\cos }^{2}}\alpha }=\sqrt{\dfrac{{{x}^{2}}-144}{{{x}^{2}}}}$
Solving this we will get
${{\cos }^{{}}}\alpha =\dfrac{\sqrt{{{x}^{2}}-144}}{{{x}^{{}}}}$
We can also write it as
$\alpha ={{\cos }^{-1}}\left( \dfrac{\sqrt{{{x}^{2}}-144}}{{{x}^{{}}}} \right)$
We have taken $\alpha ={{\sin }^{-1}}\dfrac{12}{x}$
Now we can write it as
$\alpha ={{\sin }^{-1}}\dfrac{12}{x}={{\cos }^{-1}}\left( \dfrac{\sqrt{{{x}^{2}}-144}}{{{x}^{{}}}} \right)$
Now we will use this in our given equation
${{\sin }^{-1}}\dfrac{5}{x}+{{\sin }^{-1}}\dfrac{12}{x}=\dfrac{\pi }{2}$
${{\sin }^{-1}}\dfrac{5}{x}+{{\cos }^{-1}}\left( \dfrac{\sqrt{{{x}^{2}}-144}}{{{x}^{{}}}} \right)=\dfrac{\pi }{2}$
We can write it as
${{\sin }^{-1}}\dfrac{5}{x}=\dfrac{\pi }{2}-{{\cos }^{-1}}\left( \dfrac{\sqrt{{{x}^{2}}-144}}{{{x}^{{}}}} \right)$
Taking $\sin $ both side we will get
$\sin \left( {{\sin }^{-1}}\dfrac{5}{x} \right)=\sin \left( \dfrac{\pi }{2}-{{\cos }^{-1}}\left( \dfrac{\sqrt{{{x}^{2}}-144}}{{{x}^{{}}}} \right) \right)$
Solving this we will get
$\dfrac{5}{x}=\sin \left( \dfrac{\pi }{2}-{{\cos }^{-1}}\left( \dfrac{\sqrt{{{x}^{2}}-144}}{{{x}^{{}}}} \right) \right)$
As we know $\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta $
Using this we will get
$\dfrac{5}{x}=\cos \left( {{\cos }^{-1}}\left( \dfrac{\sqrt{{{x}^{2}}-144}}{{{x}^{{}}}} \right) \right)$
Simplifying this we will get
$\dfrac{5}{x}=\left( \dfrac{\sqrt{{{x}^{2}}-144}}{{{x}^{{}}}} \right)$
Now we will solve this
By Cross multiplying we get
$5x=\left( \sqrt{{{x}^{2}}-144} \right)x$
We can write it as
$5=\left( \sqrt{{{x}^{2}}-144} \right)$
Squaring both sides we will get
${{(5)}^{2}}={{\left( \sqrt{{{x}^{2}}-144} \right)}^{2}}$
$\Rightarrow 25={{x}^{2}}-144$
Transposing $144$ to left hand side
$25+144={{x}^{2}}$
Or
$\begin{align}
  & {{x}^{2}}=25+144 \\
 & \Rightarrow {{x}^{2}}=169 \\
\end{align}$
Now we will take square root both sides
$\sqrt{{{x}^{2}}}=\sqrt{169}$
We will get
$x=\pm 13$
$\therefore x=\pm 13$ Is our required answer.
Hence $x=\pm 13$is solution of ${{\sin }^{-1}}\dfrac{5}{x}+{{\sin }^{-1}}\dfrac{12}{x}=\dfrac{\pi }{2}$

Note:
In mathematical functions we give input and get output. Input values are considered as the domain of the function and output values are considered as range of a function. In inverse functions domain of a function becomes range of an inverse function and range of function becomes domain of an inverse function.