Question

Sin alpha=3/5 and sin beta=5/13 find tan(alpha+beta)

Answer 1

Consider sin(α) = 3/5

sin²(α) = 9/25

We know,

sin²(θ) + cos²(θ) = 1

So,

sin²(α) + cos²(α) = 1

=> 9/25 + cos²(α) = 1

=> cos²(α) = 1 - (9/25)

=> cos²(α) = 16/25

=> cos(α) = 4/5

tan (α) = sin(α) / cos(α)

tan (α) = (3/5) / (4/5)

tan (α) = 3/4

Consider sin(β) = 5/13

sin²(β) = 25/169

sin²(β) + cos²(β) = 1

=> 25/169 + cos²(β) = 1

=> cos²(β) = 1 - (25/169)

=> cos²(β) = 144/169

=> cos(β) = 12/13

tan (β) = sin(β) / cos(β)

tan (β) = (5/13) / (12/13)

tan (β) = 5/12

Now,

tan (α + β) = [tan (α) + tan (β)] / [1 - tan (α) · tan (β)]

tan (α + β) = [(3/4) + (5/12)] / [1 - (3/4 · 5/12)]

tan (α + β) = (7/6) / (1 - (15/48))

tan (α + β) = (7/6) / (33/48)

tan (α + β) = (7/6) · (48/33)

tan (α + β) = 56/33

Answer 2

Answer:

jjhjhv bhbbbbbbbbbhhjoovog9h9g9h9h9h