Answer:
[tex]tan165^0 = -2 + \sqrt{3}[/tex]
Step-by-step explanation:
Given;
tan165⁰
165⁰ is in the second quadrant, and sine is positive in this quadrant, thus tan will be negative.
In second quadrant, tanθ = -tan(180 - θ)
tan165⁰ = -tan(180-165)
tan165⁰ = -tan15⁰
[tex]- tan15^0 = -tan(60 - 45)\\\\ tan(\alpha - \beta) = \frac{tan \alpha -\ tan \beta}{1 + tan \alpha tan \beta}\\\\ -tan(60 - 45) = -(\frac{tan 60 - tan 45}{1+tan60tan45} )\\\\ -(\frac{tan 60 - tan 45}{1+tan60tan45} )= -(\frac{\sqrt{3} \ - 1}{1+ \ \sqrt{3} } ) = \frac{1 \ -\sqrt{3} }{1+ \ \sqrt{3} }\\\\\frac{1 \ -\sqrt{3} }{1+ \ \sqrt{3} } =( \frac{1 \ -\sqrt{3} }{1+ \ \sqrt{3} })(\frac{1 - \ \sqrt{3}}{1 - \ \sqrt{3}} ) = \frac{1-\sqrt{3} -\sqrt{3} \ +\ 3 }{1\ -\ 3} = \frac{4-2\sqrt{3} }{-2}[/tex]
[tex]\frac{4-2\sqrt{3} }{-2} = -2 + \sqrt{3}[/tex]
Therefore, [tex]tan165^0 = -2 + \sqrt{3}[/tex]
Answer:
- sqrt(7-4sqrt(3)) = -0.26794
tan165º = -0.26794
Step-by-step explanation:
On EDG
