Question

CosA=3/5 and sinB=5/13 then the value of tanA+tanB/1-tanAtanB

Answer 1

Answer:

Step-by-step explanation:

Here,

CosA = 3/5

SinB = 5/13

Firstly,

CosA = 3/5

(cosA)^2 = (3/5)^2

(CosA)^2= 9/25

1 - (SinA)^2 = 9/25       [(cosA)^2 = 1 - (sinA)^2]

1 - 9/25 = (SinA)^2

16/25 = (SinA)^2

SinA = 4/5

Again,

SinB = 5/13

(SinB)^2 = 25/169

1 - (CosB)^2 = 25/169     [(SinB)62 = 1 - (CosB)^2]

1 - 25/169 = (CosB)^2

144/169 = (CosB)^2

12/13 = CosB

We konw,

tanA = sinA/cosA

Therefore,

tanA = SinA/CosA  ,    tanB = SinB/CosB

         = 4/3              ,             = 5/12

So,

(tanA+tanB)/(1-tanAtanB)

= (4/3 + 5/12)/(1- 4/3*5/12)

= (7/4)/(1-5/9)

=(7/4)/(4/9)

=7/9      (ANSWER)