if sin(a+b)=1 and sin(a-b)=1/2 , where 0 =a, b = /2 , then find the values of tan(a+2b) and tan(2a+b) - Maths - Trigonometric Functions

Question

if sin(a+b)=1 and sin(a-b)=1/2 , where 0<=a, b<=
π /2 , then find the values of
tan(a+2b) and tan(2a+b)

Amarsen · Accepted Answer

sin(a+b) = 1 = sin90
a + b = 90 (i)
sin(a-b) = 1/2 = sin30
a - b = 30 (ii)
adding (i) and (ii) we get
2a = 120
a = 60
by putting the value of a in any equation we get b = 30
tan(a + 2b) = tan(60 + 60) = tan(90 + 30) = - cot30 = -underroot
3
tan(2a+b) = tan(120 + 30) = tan(90+ 60) = - cot60 = -underroot
1/3

if sin(a+b)=1 and sin(a-b)=1/2 , where 0<=a, b<= π /2 , then find the values of tan(a+2b) and tan(2a+b).