Question

If we have an expression as $p\sin x=q$, then $\sqrt{{{p}^{2}}-{{q}^{2}}}\tan x$ is equal to:
A. $p$
B. $q$
C. $pq$
D. $p+q$

Answer 1

Hint: Here, we have been given that psinx=q and we have been asked to find the value of $\sqrt{{{p}^{2}}-{{q}^{2}}}\tan x$. For this, we will first square the given equation, i.e. $p\sin x=q$ on both sides. Then we will write ${{\sin }^{2}}x$ thus obtained in the equation as $1-{{\cos }^{2}}x$. Then we will separate the terms with the trigonometric function and without it on different sides of the equal to sign. Basically, we will try to bring ${{p}^{2}}-{{q}^{2}}$ on one side of the equal to sign. Once we have obtained that, we will take the square root of the equation on both sides, and then we will multiply the equation with $\tan x$. As a result, we will get $\sqrt{{{p}^{2}}-{{q}^{2}}}\tan x$ on one side of the equal to sign and then we will resolve the other side as much as possible and hence we will get our answer.

Complete step-by-step solution:
Here, we have been given that psinx=q and we need to find the value of $\sqrt{{{p}^{2}}-{{q}^{2}}}\tan x$.
For this, we will first square the given equation, i.e. $p\sin x=q$
Thus, squaring the equation on both sides, we get:
$\begin{align}
  & p\sin x=q \\
 & \Rightarrow {{\left( p\sin x \right)}^{2}}={{\left( q \right)}^{2}} \\
 & \Rightarrow {{p}^{2}}{{\sin }^{2}}x={{q}^{2}} \\
\end{align}$
Now, we know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. From this, we can also say that:
${{\sin }^{2}}x=1-{{\cos }^{2}}x$
Thus, putting the value of ${{\sin }^{2}}x$ in the now obtained equation, we get:
$\begin{align}
  & {{p}^{2}}{{\sin }^{2}}x={{q}^{2}} \\
 & \Rightarrow {{p}^{2}}\left( 1-{{\cos }^{2}}x \right)={{q}^{2}} \\
 & \Rightarrow {{p}^{2}}-{{p}^{2}}{{\cos }^{2}}x={{q}^{2}} \\
\end{align}$
Now, collecting the terms without trigonometric function in them on side of the equal to sign and the one with the trigonometric function on the other side of the sign, we get:
$\begin{align}
  & {{p}^{2}}-{{p}^{2}}{{\cos }^{2}}x={{q}^{2}} \\
 & \Rightarrow {{p}^{2}}-{{q}^{2}}={{p}^{2}}{{\cos }^{2}}x \\
\end{align}$
Now, we will take the square root on both sides of the equation.
Taking the square root of the now obtained equation, we get:
$\begin{align}
  & {{p}^{2}}-{{q}^{2}}={{p}^{2}}{{\cos }^{2}}x \\
 & \Rightarrow \sqrt{{{p}^{2}}-{{q}^{2}}}=p\cos x \\
\end{align}$
Now, we will multiply the equation with ‘$\tan x$’ on both sides.
Thus, multiplying tanx on both sides of the now obtained equation, we get:
$\begin{align}
  & \sqrt{{{p}^{2}}-{{q}^{2}}}=p\cos x \\
 & \Rightarrow \sqrt{{{p}^{2}}-{{q}^{2}}}\tan x=p\cos x.\tan x \\
\end{align}$
Now, we know that $\tan x=\dfrac{\sin x}{\cos x}$. Thus, putting this value of tanx on the RHS of the now obtained equation, we get:
$\begin{align}
  & \sqrt{{{p}^{2}}-{{q}^{2}}}\tan x=p\cos x.\tan x \\
 & \Rightarrow \sqrt{{{p}^{2}}-{{q}^{2}}}\tan x=p\cos x.\dfrac{\sin x}{\cos x} \\
 & \Rightarrow \sqrt{{{p}^{2}}-{{q}^{2}}}\tan x=p\sin x \\
\end{align}$
Now, as given in the question, $p \sin x=q$.
Thus, putting the value of psinx in the now obtained equation we get:
$\begin{align}
  & \sqrt{{{p}^{2}}-{{q}^{2}}}\tan x=p\sin x \\
 & \therefore \sqrt{{{p}^{2}}-{{q}^{2}}}\tan x=q \\
\end{align}$
Now, we can see that the LHS of this equation is the quantity we were asked to find in the question.
Thus, we get the value of $\sqrt{{{p}^{2}}-{{q}^{2}}}\tan x$ as ‘q’.
Hence, option (B) is the correct option.

Note: We here have used basic trigonometric properties which are very useful. Some of them are given as follows:
1. ${{\sin }^{2}}x+{{\cos }^{2}}x=1$
2. ${{\sec }^{2}}x=1+{{\tan }^{2}}x$
3. $\text{cose}{{\text{c}}^{2}}x=1+{{\cot }^{2}}$
4. $\tan x=\dfrac{\sin x}{\cos x}$
5. $\cot x=\dfrac{\cos x}{\sin x}$
Here we could also solve this problem by substituting the value of x in terms of $\sin x$ and $\cos x$ and again solve $\cos x$ in $\sin x$ term as we are already given its value.