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If $\cos \dfrac{{12}}{{13}}$, $\sin < 0$, how do you find tan in simplest form?

seo-qna
Last updated date: 21st Jun 2024
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Answer
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Hint: Sin is comparable to the side inverse a given point in a correct triangle to the hypotenuse. Cos is identical to the proportion of the side nearby an intense point in a right-calculated triangle to the hypotenuse.

Complete step by step answer:
Use the definition of cosine to find the known sides of the unit circle right triangle. The quadrant determines the sign on each of the values.
$\cos (x) = \dfrac{{adjacent}}{{hypotenuse}}$
Let's find the opposite side of the unit circle triangle. Since the adjacent side and hypotenuse are known, use the Pythagorean theorem to find the remaining side.
$opposite = \sqrt {hypotenuse{e^2} - adjacent{t^2}} $
Replace the known value of the in the equation.
$opposite = \sqrt {{{13}^2} - {{12}^2}} $
Simplify $\sqrt {{{13}^2} - {{12}^2}} $
Raise 13 to the power of 2.
$Opposite = \sqrt {169 - {{(12)}^2}} $
Raise 12 to the power of 2.
$Opposite = \sqrt {169 - 1.144} $
Multiply $ - 1$ by 144
$Opposite = \sqrt {169 - 144} $
Subtract 144 from 169.
$Opposite = \sqrt {25} $
Rewrite 25 as ${5^2}$.
$Opposite = \sqrt {{5^2}} $
Pull terms out from under the radical, assuming positive real numbers.
$Opposite = 5$
Use the definition of $\sin $to find the value of $\sin (x)$.
$\sin (x) = \dfrac{{opp}}{{hyp}}$
Substitute in the known values.
$\sin (a) = \dfrac{5}{{13}}$.
$cos(a) = \dfrac{{12}}{{13}}$. Angle is in either 1st quadrant or in the 4th.
$\sin (a) = \pm \dfrac{5}{{13}}$. As, $\sin (a) < 0$, a is in the 4th quadrant. So, $\sin (a) = - \dfrac{5}{{13}}$.
Use the definition of tangent to find the value of $\tan (x)$
$\tan (x) = \dfrac{{opp}}{{adj}}$
Substitute in the known values.
$\tan (a) = - \dfrac{5}{{12}}$.

Thus, the ratio of $\dfrac{{\sin (a)}}{{\cos (a)}} = \tan (a) = - \dfrac{5}{{12}}$.

Note: We note that the domain of sin inverse function is $( - 1,1)$ and since $\dfrac{{12}}{{13}} \in ( - 1,1)$ the value $\sin = - \dfrac{5}{{13}}$ is well defined.