Question

If cos⁡ x=−45, and 180° < x < 270°, what is tan⁡(2x)?

Answer 1

$Tan2x =\frac{-24}{7}$

Step-by-step explanation:

Here we have , cos x = -4/5 ( corrected as given -45 which is not possible ) ,

180° < x < 270° i.e. in third quadrant  . We need to find tan⁡(2x) . Let's find out:

⇒ $Tan\alpha = \frac{sin\alpha }{cos\alpha }$

⇒ $Tan\alpha = \frac{\sqrt{1-(cos\alpha)^2} }{cos\alpha }$

⇒ $Tan\alpha = \frac{\sqrt{1-(\frac{-4}{5})^2} }{\frac{-4}{5} }$

⇒ $Tan\alpha = \frac{-5}{4}\sqrt{1-(\frac{16}{25}) }$

⇒ $Tan\alpha = \frac{-5}{4}\sqrt{(\frac{25-16}{25}) }$

⇒ $Tan\alpha = \frac{-5}{4}(\frac{3}{5}) }$

⇒ $Tan\alpha = \frac{-3}{4}$ , since in third quadrant . $Tan\alpha = \frac{-3}{4}$ remains same .

Now , $Tan2x = \frac{2tanx}{1-(tanx)^{2}}}$

⇒ $Tan2x = \frac{2tanx}{1-(tanx)^{2}}}$

⇒ $Tan2x = \frac{2(\frac{-3}{4})}{1-(\frac{-3}{4})^{2}}}$

⇒ $Tan2x = \frac{-3}{2}(\frac{16}{7})$

⇒ $Tan2x =\frac{-24}{7}$

Therefore, $Tan2x =\frac{-24}{7}$ .