Question

How do you solve ${\cot ^2}x{\csc ^2}x - {\cot ^2}x = 9\;\;\;{\text{for}}\;\;0 \leqslant x\leqslant 2\pi$?

Answer 1

Now in order to solve the above equation we should work with one side at a time and manipulate it to the other side. Using some basic trigonometric identities like we can simplify the above expression.
$\cot x = \dfrac{1}{{\tan x}} \\\ 1 + {\cot ^2}x = {\csc ^2}x \\\$
In order to solve the given expression we have to use the above identities and express our given expression in that form and thereby solve it.

Complete step by step solution:
Given

............................\left( i \right)$$ Here we have to basically solve for $x$ using any mathematical operations and functions. In order to solve for $x$ we can use the above mentioned trigonometric identities to simplify it. Now we can see that the term ${\cot ^2}x$ is common to both the terms: So taking the term ${\cot ^2}x$ , we can write:

{\cot ^2}x{\csc ^2}x - {\cot ^2}x = 9; \\
{\cot ^2}x\left( {{{\csc }^2}x - 1} \right) = 9............................\left( {ii} \right) \\

Now we know the equation $ 1 + {\cot ^2}x = {\csc ^2}x \\\ \Rightarrow {\csc ^2}x - 1 = {\cot ^2}x \\\ $ So on substituting it in (ii) we can write:

{\cot ^2}x\left( {{{\csc }^2}x - 1} \right) = 9 \\
{\cot ^2}x\left( {{{\cot }^2}x} \right) = 9....................\left( {iii} \right) \\

Taking square root on both sides of the equation (iii) we can write: $${\cot ^2}x = 3........................\left( {iv} \right)$$ Once again taking square root on both sides of the equation (iv) we can write

{\cot ^2}x = 3 \\
\Rightarrow \cot x = \pm \sqrt 3 \\

Now we know that: $\cot x = \dfrac{1}{{\tan x}}$ Such that the above equation we can write:

\dfrac{1}{{\tan x}} = \pm \sqrt 3 \\
\tan x = \pm \dfrac{1}{{\sqrt 3 }} \\
\Rightarrow x = n\pi + \dfrac{\pi }{6} \\

Now we have to find $x$ $${\text{for}}\;\;0 \leqslant x \leqslant 2\pi $$. **So by solving $${\cot ^2}x{\csc ^2}x - {\cot ^2}x = 9\;\;\;{\text{for}}\;\;0 \leqslant x \leqslant 2\pi $$ we get $x = \dfrac{\pi }{6},7\dfrac{\pi }{6}.$** **Note:** Some other equations needed for solving these types of problem are:

1 + {\tan ^2}x = {\sec ^2}x \\
\begin{array}{*{20}{l}}
{\sin \left( {2x} \right) = 2\sin \left( x \right)\cos \left( x \right)} \\
{\cos \left( {2x} \right) = {{\cos }^2}\left( x \right)-{{\sin }^2}\left( x \right) = 1-2{\text{ }}{{\sin
}^2}\left( x \right) = 2{\text{ }}{{\cos }^2}\left( x \right)-1}
\end{array} \\

$$Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.$$