Question

If x = nπ - ${\text{ta}}{{\text{n}}^{ - 1}}$3 is a solution of the equation 12tan2x + $\dfrac{{\sqrt {10} }}{{{\text{cos x}}}}$+ 1 = 0 then
$
  {\text{A}}{\text{. n is any integer}} \\
  {\text{B}}{\text{. n is an even integer}} \\
  {\text{C}}{\text{. n is a positive integer}} \\
  {\text{D}}{\text{. n is an odd integer}} \\
$

Answer 1

Hint – To compute the given equation, we substitute the given value of x in the equation and solve it for n is even and n is odd and verify which comes out to be true. Therefore we determine the property of n.
Complete Step-by-Step solution:
Given Data, 12tan2x + $\dfrac{{\sqrt {10} }}{{{\text{cos x}}}}$+ 1
We put x = nπ - ${\text{ta}}{{\text{n}}^{ - 1}}$3 in this equation we get,
$ \Rightarrow 12{\text{ tan }}\left( {{\text{2n}}\pi {\text{ - 2ta}}{{\text{n}}^{ - 1}}3} \right) + \dfrac{{\sqrt {10} }}{{{\text{cos }}\left( {{\text{n}}\pi {\text{ - ta}}{{\text{n}}^{ - 1}}3} \right)}} + 1{\text{ = 0}}$

Now let n be even,
$ \Rightarrow 12{\text{ tan }}\left( {{\text{2n}}\pi {\text{ - 2ta}}{{\text{n}}^{ - 1}}3} \right) + \dfrac{{\sqrt {10} }}{{{\text{cos }}\left( {{\text{n}}\pi {\text{ - ta}}{{\text{n}}^{ - 1}}3} \right)}} + 1{\text{ = 0}}$
We know tan (nπ – θ) = -tan θ and cos (nπ – θ) = cos θ, when n is an even number
$ \Rightarrow - 12{\text{ tan }}\left( {{\text{2ta}}{{\text{n}}^{ - 1}}3} \right) + \dfrac{{\sqrt {10} }}{{{\text{cos }}\left( {{\text{ta}}{{\text{n}}^{ - 1}}3} \right)}} + 1{\text{ = 0}}$
Now we know, ${\tan ^{ - 1}}3{\text{ = 71}}{\text{.56}}^\circ $and cos 71.56° =$\dfrac{1}{{\sqrt {10} }}$, also tan (2 × 71.56°) = $ - \dfrac{3}{4}$
We get these values of tan, tan inverse and cos functions from their respective trigonometric tables.
$
   \Rightarrow - 12\left( { - \dfrac{3}{4}} \right) + \dfrac{{\sqrt {10} }}{{\left( {\dfrac{1}{{\sqrt {10} }}} \right)}} + 1{\text{ = 0}} \\
   \Rightarrow {\text{9 + 10 + 1 = 0}} \\
   \Rightarrow {\text{20}} \ne 0 \\
$
Hence n is not an even integer.

Now let n be an odd number,
$ \Rightarrow 12{\text{ tan }}\left( {{\text{2n}}\pi {\text{ - 2ta}}{{\text{n}}^{ - 1}}3} \right) + \dfrac{{\sqrt {10} }}{{{\text{cos }}\left( {{\text{n}}\pi {\text{ - ta}}{{\text{n}}^{ - 1}}3} \right)}} + 1{\text{ = 0}}$
We know tan (nπ – θ) = tan θ and cos (nπ – θ) = -cos θ, when n is an odd number
$ \Rightarrow 12{\text{ tan }}\left( {{\text{2ta}}{{\text{n}}^{ - 1}}3} \right) + \dfrac{{\sqrt {10} }}{{{\text{ - cos }}\left( {{\text{ta}}{{\text{n}}^{ - 1}}3} \right)}} + 1{\text{ = 0}}$
Now we know, ${\tan ^{ - 1}}3{\text{ = 71}}{\text{.56}}^\circ $and cos 71.56° =$\dfrac{1}{{\sqrt {10} }}$, also tan (2 × 71.56°) = $ - \dfrac{3}{4}$
$
   \Rightarrow - 12\left( { - \dfrac{3}{4}} \right) - \dfrac{{\sqrt {10} }}{{\left( {\dfrac{1}{{\sqrt {10} }}} \right)}} + 1{\text{ = 0}} \\
   \Rightarrow {\text{9 - 10 + 1 = 0}} \\
   \Rightarrow {\text{0 = }}0 \\
$
Hence n is an odd integer.
Option D is the correct answer.

Note – In order to solve this type of problem the key is to verify the possible values of n one by one. It is important to have adequate knowledge in using the trigonometric tables of tan and cos functions, also their inverse functions. WE have to be careful while converting the angles inside the tan and cos functions,
tan (nπ – θ) = -tan θ and cos (nπ – θ) = cos θ, when n is an even number.
tan (nπ – θ) = tan θ and cos (nπ – θ) = -cos θ, when n is an odd number.