Question

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

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 $\tan {\text{ and }}\cot$.

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.