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\[\dfrac{{\cos {{10}^0} + \sin {{10}^0}}}{{\cos {{10}^0} - \sin {{10}^0}}}\] is equal to
A. \[\tan {55^0}\]
B. \[\cot {55^0}\]
C. \[ - \tan {35^0}\]
D. \[ - \cot {35^0}\]

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Last updated date: 20th Jun 2024
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Answer
VerifiedVerified
54.9k+ views
Hint: In this problem just multiply with the suitable trigonometric ratio and convert the given equation in terms of \[\tan {\text{ or }}\cot \] by using the simple trigonometric formulae since the given options are in terms of \[\tan {\text{ and }}\cot \].

Complete step-by-step answer:
Given \[\dfrac{{\cos {{10}^0} + \sin {{10}^0}}}{{\cos {{10}^0} - \sin {{10}^0}}}\]
Multiplying and dividing with \[\cos {10^0}\] then we have
\[
   \Rightarrow \dfrac{{\cos {{10}^0}}}{{\cos {{10}^0}}}\left( {\dfrac{{\cos {{10}^0} + \sin {{10}^0}}}{{\cos {{10}^0} - \sin {{10}^0}}}} \right) \\
    \\
  \dfrac{{ \Rightarrow \dfrac{{\cos {{10}^0}}}{{\cos {{10}^0}}} + \dfrac{{\sin {{10}^0}}}{{\cos {{10}^0}}}}}{{\dfrac{{\cos {{10}^0}}}{{\cos {{10}^0}}} - \dfrac{{\sin {{10}^0}}}{{\cos {{10}^0}}}}} \\
\]

Since \[\dfrac{{\sin {{10}^0}}}{{\cos {{10}^0}}} = \tan {10^0}\]

\[ \Rightarrow \dfrac{{1 + \tan {{10}^0}}}{{1 - \tan {{10}^0}}}\]
We can write \[\tan {45^0}\]in place of \[1\] as \[\tan {45^0} = 1\] then we get
\[ \Rightarrow \dfrac{{\tan {{45}^0} + \tan {{10}^0}}}{{1 - \tan {{45}^0}\tan {{10}^0}}}\]
By using the formulae \[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\] we have
\[
   \Rightarrow \dfrac{{\tan {{45}^0} + \tan {{10}^0}}}{{1 - \tan {{45}^0}\tan {{10}^0}}} = \tan \left( {{{45}^0} + {{10}^0}} \right) \\
    \\
  {\text{ = tan5}}{{\text{5}}^0} \\
\]

Thus, \[\dfrac{{\cos {{10}^0} + \sin {{10}^0}}}{{\cos {{10}^0} - \sin {{10}^0}}}\] is equal to \[\tan {55^0}\]

Therefore, the answer is option A \[\tan {55^0}\]

Note: In this problem there are chances to change the options by converting \[\tan \]into \[\cot \]or from\[\tan \] to \[\cot \]. Then we have to change them accordingly. And try to remember more formulae from the trigonometry part so that you can make problems easier.