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  • The value of  <img alt="\frac{cos\ 20^{\circ}+sin\ 20^{\circ}}{cos\ 20^{\circ}-sin\ 20^{\circ}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7Bcos%5C%2020%5E%7B%5Ccirc%7D

The value of  \frac{cos\ 20^{\circ}+sin\ 20^{\circ}}{cos\ 20^{\circ}-sin\ 20^{\circ}} is 

Option: 1

\cot\ 65^{\circ}


Option: 2

\tan\ 65^{\circ}


Option: 3

-\cot\ 65^{\circ}


Option: 4

-\tan\ 65^{\circ}


Answers (1)

best_answer

Given that , 

\frac{cos\ 20^{\circ}+sin\ 20^{\circ}}{cos\ 20^{\circ}-sin\ 20^{\circ}}

Dividing the numerator and denominator by \cos\ 20^{\circ},

\frac{cos\ 20^{\circ}+sin\ 20^{\circ}}{cos\ 20^{\circ}-sin\ 20^{\circ}}=\frac{\frac{cos\ 20^{\circ}+sin\ 20^{\circ}}{cos\ 20^{\circ}}}{\frac{cos\ 20^{\circ}-sin\ 20^{\circ}}{{cos\ 20^{\circ}}}}

Substituting the formulas of trigonometric ratios of compound angles,

\\\frac{cos\ 20^{\circ}+sin\ 20^{\circ}}{cos\ 20^{\circ}-sin\ 20^{\circ}}=\frac{1+tan\ 20^{\circ}}{1-tan\ 20^{\circ}}\\ \\\\\frac{cos\ 20^{\circ}+sin\ 20^{\circ}}{cos\ 20^{\circ}-sin\ 20^{\circ}}=\frac{tan\ \frac{\pi }{4}+tan\ 20^{\circ}}{1-tan\ \frac{\pi }{4}\ tan\ 20^{\circ}}

\\\frac{cos\ 20^{\circ}+sin\ 20^{\circ}}{cos\ 20^{\circ}-sin\ 20^{\circ}}=tan\left ( \frac{\pi }{4}+20^{\circ} \right )\\ \\\\\frac{cos\ 20^{\circ}+sin\ 20^{\circ}}{cos\ 20^{\circ}-sin\ 20^{\circ}}=tan\left ( 45+20 \right )

\frac{cos\ 20^{\circ}+sin\ 20^{\circ}}{cos\ 20^{\circ}-sin\ 20^{\circ}}=tan\ 65^{\circ}

Posted by

shivangi.bhatnagar

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