If the eccentricities of the hyperbolas \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\] and \[\dfrac{{{y^2}}}{{{b^2}}} - \dfrac{{{x^2}}}{{{a^2}}} = 1\] be \[e\] and \[{e_1}\]. Then what is the value of \[\dfrac{1}{{{e^2}}} + \dfrac{1}{{{e_1}^2}}\]?
A. 1
B. 2
C. 3
D. None of these

Hint: Here, the equations of the two hyperbolas are given. First, apply the formula of the eccentricity of the hyperbola and calculate the eccentricities of both hyperbolas. Then, substitute the values in the given expression and solve it to get the required answer.

Formula used: The eccentricity of the hyperbola \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\] is: \[e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} \]
The eccentricity of the hyperbola \[\dfrac{{{y^2}}}{{{b^2}}} - \dfrac{{{x^2}}}{{{a^2}}} = 1\] is: \[e = \sqrt {1 + \dfrac{{{a^2}}}{{{b^2}}}} \]

Complete step by step solution: Given:
The equations of the hyperbolas are \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\] and \[\dfrac{{{y^2}}}{{{b^2}}} - \dfrac{{{x^2}}}{{{a^2}}} = 1\].
\[e\] and \[{e_1}\] are the eccentricities of both hyperbolas respectively.

Let’s calculate the eccentricities of the both hyperbolas.
From the standard formulas of the eccentricity of the hyperbola, we get
The eccentricity of the hyperbola \[\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1\] is:
\[e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} \]
\[ \Rightarrow e = \sqrt {\dfrac{{{a^2} + {b^2}}}{{{a^2}}}} \]
Take squares of both sides.
\[ \Rightarrow {e^2} = \dfrac{{{a^2} + {b^2}}}{{{a^2}}}\] \[.....\left( 1 \right)\]

Also, the eccentricity of the hyperbola \[\dfrac{{{y^2}}}{{{b^2}}} - \dfrac{{{x^2}}}{{{a^2}}} = 1\] is:
\[{e_1} = \sqrt {1 + \dfrac{{{a^2}}}{{{b^2}}}} \]
\[ \Rightarrow {e_1} = \sqrt {\dfrac{{{b^2} + {a^2}}}{{{b^2}}}} \]
Take squares of both sides.
\[ \Rightarrow {e_1}^2 = \dfrac{{{a^2} + {b^2}}}{{{b^2}}}\] \[.....\left( 2 \right)\]

Now substitute the equations \[\left( 1 \right)\] and \[\left( 2 \right)\] in the required equation.
\[\dfrac{1}{{{e^2}}} + \dfrac{1}{{{e_1}^2}} = \dfrac{1}{{\dfrac{{{a^2} + {b^2}}}{{{a^2}}}}} + \dfrac{1}{{\dfrac{{{a^2} + {b^2}}}{{{b^2}}}}}\]
\[ \Rightarrow \dfrac{1}{{{e^2}}} + \dfrac{1}{{{e_1}^2}} = \dfrac{{{a^2}}}{{{a^2} + {b^2}}} + \dfrac{{{b^2}}}{{{a^2} + {b^2}}}\]
\[ \Rightarrow \dfrac{1}{{{e^2}}} + \dfrac{1}{{{e_1}^2}} = \dfrac{{{a^2} + {b^2}}}{{{a^2} + {b^2}}}\]
\[ \Rightarrow \dfrac{1}{{{e^2}}} + \dfrac{1}{{{e_1}^2}} = 1\]

Thus, Option (B) is correct.

Note: Sometimes students get confused and write the eccentricity of the hyperbola \[\dfrac{{{y^2}}}{{{b^2}}} - \dfrac{{{x^2}}}{{{a^2}}} = 1\] as \[e = \sqrt {1 + \dfrac{{{b^2}}}{{{a^2}}}} \]. Because of that, they get the wrong answer.