If sin theta is equal to 2/3, theta not in quadrant 1, find tan theta?

Question

Quinn Anderson · Accepted Answer

#tantheta=-2/sqrt5# #"using the "color(blue)"trigonometric identities"# #•color(white)(x)sin^2theta+cos^2theta=1# #rArrcostheta=+-sqrt(1-sin^2theta)# #•color(white)(x)tantheta=sintheta/costheta# #"since "sintheta>0" but not in first quadrant then"# #theta" is in second quadrant"# #"where "costheta" and "tantheta<0# #costheta=-sqrt(1-(2/3)^2)# #color(white)(costheta)=-sqrt(1-4/9)=-sqrt(5/9)=-sqrt5/3# #tantheta=(2/3)/(-sqrt5/3)=2/3xx-3/sqrt5=-2/sqrt5#

Charles Autterson · Answer

#tantheta=-2/sqrt5#

Here,

#sintheta=2/3 > 0=>I^(st)Quadrant or II^(nd)Quadrant#

#(i) #But given that , theta is not in quadrant 1

So, we have

#(ii)pi/2 < theta < pi toII^(nd)Quadrant#

#:.sintheta=2/3 >0 , costheta < 0and tantheta < 0#

Now , using trig.identity , we get

#cos^2theta=1-sin^2theta=1-4/9=5/9=(sqrt5/3)^2#

#=>costheta=-sqrt5/3 < 0#

Hence,

#tantheta=sintheta/costheta=(2/3)/(-sqrt5/3)=-2/sqrt5 < 0#

Grace Ashcraft · Answer

undefined Given that $ \sin(\theta) = \frac{2}{3} $ and $ \theta $ is not in quadrant 1, we can use the relationship between sine and cosine in different quadrants to determine the sign of $ \cos(\theta) $. Since $ \sin(\theta) = \frac{2}{3} $ is positive, $ \cos(\theta) $ must also be positive because in quadrants where sine is positive, cosine is also positive.

Using the Pythagorean identity $ \sin^2(\theta) + \cos^2(\theta) = 1 $, we can solve for $ \cos(\theta) $ to find $ \cos(\theta) = \sqrt{1 - \sin^2(\theta)} $. Then, we can find $ \tan(\theta) $ using the relationship $ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $.

Substituting the given values, we have $ \tan(\theta) = \frac{\frac{2}{3}}{\sqrt{1 - \left(\frac{2}{3}\right)^2}} $. After simplification, we get $ \tan(\theta) = \frac{2}{\sqrt{5}} $. Therefore, $ \tan(\theta) = \frac{2\sqrt{5}}{5} $ because $ \sqrt{5} \times \sqrt{5} = 5 $.