Question

Find tanA if cosA=-12/13 and A lies in the third quadrant.

Answer 1

Answer:

$\frac{5}{12}$

Step-by-step explanation:

A lies in third quadrant

tan A is positive in third quadrant while sin A is negative in third quadrant.

Now,

${ \sin(a) }^{2} + { \cos(a) }^{2} = 1 \\ { \sin(a) }^{2} + ( { - \frac{ 12}{13} ) }^{2} = 1 \\ { \sin( a ) }^{2} + \frac{144}{169} = 1 \\ { \sin(a) }^{2} = \frac{169 - 144}{169} \\ { \sin(a) }^{2} = \frac{25}{169 } \\ \sin(a) = \sqrt{ \frac{25}{169} } = - \frac{5}{13}$

Then

$\tan(a) = \frac{ \sin(a) }{ \cos(a) } \\ = \frac{ - 5}{13} \div \frac{ - 12}{13} \\ = \frac{5 \times 13}{12 \times 13} = \frac{5}{12}$

5/12 is the answer.

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Answer 2

Answer:

answer is 5/12

Step-by-step explanation:

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