Given that 0 ≀ π‘₯ ≀ πœ‹/2and 0 ≀ 𝑦 β‰€πœ‹/2

find the exact value of π‘‘π‘Žπ‘›(π‘₯ + 𝑦) if π‘π‘œπ‘‘(π‘₯) =6/5 and 𝑠𝑒𝑐(𝑦) =3/2

cotx = 6/5 ---> tanx = 5/6

secy = 3/2 ---> cosy = 2/3

sketch your triangles in quad I
using Pythagoras,
if tanx = 5/6
r^2 = 25 + 36 = 61
r = √61

if cosy = 2/3
by Pythagoras: 2^2 + Y^2 = 3^2
Y = √5
tany = √5/3

tan(x+y) = (tanx + tany)/(1 - tanxtany)
= (5/6 + √5/3)/(1 - (5/6)(√5/3)
= ( (5 + 2√5)/6 )/ ( (6 - 5√5)/6)
= (5 + 2√5)/(6 - 5√5)