# csc lambda = 17/8 in quadrant 2. What is tan 2lambda?

Apr 16, 2016

First of all, we must remember that $\csc = \frac{1}{\sin}$

#### Explanation:

Thus, we can conclude that $\sin \lambda = \frac{8}{17}$

Since sin is opposite/hypotenuse, we must find the adjacent side, because tan is opposite/adjacent. We can do this by pythagorean theorem. The following diagram shows how we can draw a right triangle on the cartesian plane to find the value of the six trigonometric ratios.

Rearranging our pythagorean theorem to find adjacent side $b$, we get:

${b}^{2} = {c}^{2} - {a}^{2}$

${b}^{2} = {17}^{2} - {8}^{2}$

${b}^{2} = 225$

$b = 15$

Therefore, the adjacent side from $\lambda$ measures -15 units (the x axis is negative in quadrant II). We can conclude that $\tan \lambda = - \frac{8}{15}$

Now, since it's $2 \lambda$ that we must find, we need to use the double angle formula for $\tan$: tan(2lambda) = (2tanlamda)/(1 - tan^2lamda

Substituting:

$\tan \left(2 l a m \mathrm{da}\right) = \frac{2 \left(- \frac{8}{15}\right)}{1 - {\left(- \frac{8}{15}\right)}^{2}}$

$\tan \left(2 \lambda\right) = \frac{- \frac{16}{15}}{\frac{161}{225}}$

$t n a \left(2 l a m \mathrm{da}\right) = - \frac{16}{15} \times \frac{225}{161}$

$\tan \left(2 \lambda\right) = - \frac{240}{161}$

Hopefully this helps