Asked by Stacy
                Given that 0 ≤ 𝑥 ≤ 𝜋/2and 0 ≤ 𝑦 ≤𝜋/2
find the exact value of 𝑡𝑎𝑛(𝑥 + 𝑦) if 𝑐𝑜𝑡(𝑥) =6/5 and 𝑠𝑒𝑐(𝑦) =3/2
            
        find the exact value of 𝑡𝑎𝑛(𝑥 + 𝑦) if 𝑐𝑜𝑡(𝑥) =6/5 and 𝑠𝑒𝑐(𝑦) =3/2
Answers
                    Answered by
            Reiny
            
    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)
 
    
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)
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