Number of solutions of the equations \[{{\tan }^{-1}}\left( \dfrac{1}{2x+1} \right)+{{\tan }^{-1}}\left( \dfrac{1}{4x+1} \right)\]\[={{\tan }^{-1}}\left( \dfrac{2}{{{x}^{2}}} \right)\]
(1) \[1\]
(2) \[2\]
(3) \[3\]
(4) \[4\]

Hint: To solve this question you should know about inverse trigonometry and its formulas. To solve this, first we have to solve the equation using a suitable formula and then we will find the value of x using the middle term. If the formulas are not clear to you then it will be difficult for you to solve the problem.

Complete step-by-step solution:
The inverse trigonometric functions are the inverse of sine, cosine, tangent, secant, cosecant, and cotangent.
Also, an inverse trigonometric function is used to determine the angle when its two sides are known. Choosing the sides of a triangle depends upon the angle which you have to find.
Suppose, if you want to find the sine of any angle then the perpendicular and hypotenuse of a triangle will be involved.
If you want to find the cosine of any angle then base and hypotenuse will be involved and if you want to find the tangent of any angle then perpendicular and base will be involved which is shown below.
\[\sin \theta =\dfrac{P}{H}\]
\[\cos \theta =\dfrac{B}{H}\]
\[\tan \theta =\dfrac{P}{B}\] \[\]
If we have to find the inverse of the trigonometric function, then we are changing the values of x and y.
To solve this question we should know the formula of \[{{\tan }^{-1}}a+{{\tan }^{-1}}b\] which is equal to \[{{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)\] for \[xy<1\] and \[\pi +{{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)\] for \[xy>1\].
Now,
We are given ,\[{{\tan }^{-1}}\left( \dfrac{1}{2x+1} \right)+{{\tan }^{-1}}\left( \dfrac{1}{4x+1} \right)\] \[={{\tan }^{-1}}\left( \dfrac{2}{{{x}^{2}}} \right)\]
(On Applying the formula \[{{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)\])
we will get, \[{{\tan}^{-1}}\left(\dfrac{\dfrac{1}{1+2x}+\dfrac{1}{1+4x}}{1-\left( \dfrac{1}{1+2x}\right)\left(\dfrac{1}{1+4x}\right)}\right)\]\[={{\tan}^{-1}}\left(\dfrac{2}{{{x}^{2}}} \right)\]
Now solving the numerator and denominator,\[{{\tan }^{-1}}\left(\dfrac{\dfrac{1+4x+1+2x}{\left( 1+2x \right)\left( 1+4x \right)}}{\dfrac{\left( 1+2x \right)\left( 1+4x \right)-1}{\left( 1+2x \right)\left( 1+4x \right)}} \right)\] \[={{\tan }^{-1}}\left( \dfrac{2}{{{x}^{2}}} \right)\]
Now as the base of both numerator and denominator are the same it will cancel out and will give us the following equation
\[{{\tan }^{-1}}\left( \dfrac{2+6x}{1+6x+8{{x}^{2}}-1} \right)\] \[={{\tan }^{-1}}\left( \dfrac{2}{{{x}^{2}}} \right)\]
Now equating both the sides, tan inverse will cancel out and will give
\[\dfrac{2+6x}{8{{x}^{2}}+6x}=\dfrac{2}{{{x}^{2}}}\]
Now cross multiplying, we will get
\[6{{x}^{3}}+2{{x}^{2}}=12x+16{{x}^{2}}\]
Now , \[6{{x}^{3}}-14{{x}^{2}}-12x=0\]
Now taking \[2\] common from the equation we will get,
\[3{{x}^{3}}-7{{x}^{2}}-6x=0\]
Now taking \[x\] as common we will get
 \[x\left( 3{{x}^{2}}-7x-6 \right)\]\[=0\]
Now solving the quadratic equation by using split the middle term we will get
\[x\left( 3{{x}^{2}}-9x+2x-6 \right)\]\[=0\]
Now,
\[x\{3x(x-3)+2(x-3)\}\]\[=0\]
Now we will get,
\[x(3x+2)(x-3)=0\]
Now equating this we will get
\[x=0,x=3,x=\dfrac{-2}{3}\]
As we have seen we got \[3\]values of \[x\], so the answer to this question is \[3\].

Note: Trigonometric ratios are defined as the ratio of the sides of the right-angled triangle. When two right-angled triangles are the same then their sides are proportional and hence their ratios will be equal. Trigonometric ratios are used to determine the length of one side of a triangle when all other sides are given.