Question

If \[\dfrac{\sin x}{\cos x}\times \dfrac{\sec x}{\operatorname{cosec}x}\times \dfrac{\tan x}{\cot x}=9\], where \[x\in \left( 0,\dfrac{\pi }{2} \right)\], then find the value of x.

Answer 1

Hint: Simplify the LHS first. Use the following formulae: \[\sec x=\dfrac{1}{\cos x}\], \[cosecx=\dfrac{1}{\sin x}\], \[\tan x=\dfrac{\sin x}{\cos x}\], and \[\cot x=\dfrac{1}{\tan x}\]. Then use the fact that $\tan \dfrac{\pi }{3}=\sqrt{3}$ to arrive at the final answer.

Complete step-by-step answer:

In this question, we are given that \[\dfrac{\sin x}{\cos x}\times \dfrac{\sec x}{\operatorname{cosec}x}\times \dfrac{\tan x}{\cot x}=9\], where \[x\in \left( 0,\dfrac{\pi }{2} \right)\].
We need to find the value of $x$ which satisfies this condition.
Let us simplify the LHS.
The LHS is equal to \[\dfrac{\sin x}{\cos x}\times \dfrac{\sec x}{\operatorname{cosec}x}\times \dfrac{\tan x}{\cot x}\]
Now, we know that $sec x$ is the reciprocal of $cos x$.
i.e. \[\sec x=\dfrac{1}{\cos x}\]
Similarly, we know that $cosec x$ is the reciprocal of $sin x$.
i.e. \[cosecx=\dfrac{1}{\sin x}\]
Substituting these in the given expression, we will get the following:
\[\dfrac{\sin x}{\cos x}\times \dfrac{\tfrac{1}{\cos x}}{\tfrac{1}{\sin x}}\times \dfrac{\tan x}{\cot x}=9\]
\[\dfrac{\sin x}{\cos x}\times \dfrac{\sin x}{\cos x}\times \dfrac{\tan x}{\cot x}=9\]
\[\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}\times \dfrac{\tan x}{\cot x}=9\]
Now, we also know that $cot x$ is the reciprocal of $tan x$.
i.e. \[\cot x=\dfrac{1}{\tan x}\]
Substituting this in the above expression, we will get the following:
\[\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}\times \dfrac{\tan x}{\tfrac{1}{\tan x}}=9\]
\[\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}\times {{\tan }^{2}}x=9\]
Now, we also know that $tan x$ is equal to the quotient when $sin x$ is divided by $cos x$
i.e. \[\tan x=\dfrac{\sin x}{\cos x}\]
Substituting this in the above expression, we will get the following:
\[\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}\times \dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}=9\]
\[\dfrac{{{\sin }^{4}}x}{{{\cos }^{4}}x}=9\]
\[{{\left( \dfrac{\sin x}{\cos x} \right)}^{4}}=9\]
Taking root on both sides of the above equation, we will get the following:
\[{{\left( \dfrac{\sin x}{\cos x} \right)}^{2}}=3\]
Taking root again on both sides of the above equation, we will get the following:
\[\left( \dfrac{\sin x}{\cos x} \right)=\sqrt{3}\]
The root will not be equal to \[-\sqrt{3}\], because we are given the domain of $x$ which is \[x\in \left( 0,\dfrac{\pi }{2} \right)\] . This is the first quadrant and, in this domain, $sin x$ and $cos x$ are both positive. So, their quotient cannot be negative.
Now, since \[\tan x=\dfrac{\sin x}{\cos x}\]
So, the above equation can be written as:
$\tan x=\sqrt{3}$
So, $x=\dfrac{\pi }{3}$
This is the final answer.

Note: In this question, it is very important to keep in mind that while taking root on both sides, the root will not be equal to \[-\sqrt{3}\], because we are given the domain of x which is \[x\in \left( 0,\dfrac{\pi }{2} \right)\] . This is the first quadrant and, in this domain, $sin x$ and $cos x$ are both positive. So, their quotient cannot be negative.