Problem: If sinx+cosx=1/5 and 0β©½x<Ο, then tanx is
Answer Choices:
A. β34β
B. β43β
C. 43β
D. 34β
E. not completely determined by the given information
Solution:
If sinx+cosx=51β, then
1βsin2x=cos2x=(51ββsinx)225sin2xβ5sinxβ12=0β
Similarly, cosx satisfies the equation 25t2β5tβ12=0, whose solutions are 54β and β53β. Since sinxβ©Ύ0,sinx=54β; and sincecosx=51ββsinx, cosx=β53β. Hence, tanx=β34β.