# Thread: tanA=12/5 and A=6B then find 12cosec2B + 5sec2B

1. ## tanA=12/5 and A=6B then find 12cosec2B + 5sec2B

tanA=12/5 and A=6B then find 12cosec2B + 5sec2B

2. ## Re: tanA=12/5 and A=6B then find 12cosec2B + 5sec2B

Okay, do it! Start with a Right Triangle. Label A and B. Lable the legs so that tan(A) = 12/5. Determine the hypotenuse and you are well on your way.

3. ## Re: tanA=12/5 and A=6B then find 12cosec2B + 5sec2B

Originally Posted by TKHunny
Okay, do it! Start with a Right Triangle. Label A and B. Lable the legs so that tan(A) = 12/5. Determine the hypotenuse and you are well on your way.
but that will be tan A and lets say ill get the value of tan6B and then how will i get tan 2B

4. ## Re: tanA=12/5 and A=6B then find 12cosec2B + 5sec2B

If A= 6B then B= A/6 so 2B= A/3.

Hopefully, you know that sin(x+ y)= sin(x)cos(y)+ cos(x)sin(y) and that cos(x+y)= cos(x)cos(y)- sin(x)sin(y). Taking x= y, sin(2x)= 2sin(x)cos(x) and $cos(2x)= cos^2(x)- sin^2(x)$. Then $sin(3x)= sin(2x+ x)= sin(2x)cos(x)+ cos(2x)sin(x)= 2sin(x)cos^2(x)+ (cos^2(x)- sin^2(x))sin(x)= 3sin(x)cos^2(x)- sin^3(x)$.
$cos(3x)= cos(2x+ x)= cos(2x)cos(x)- sin(2x)sin(x)= (cos^2(x)- sin^2(x))cos(x)- 2sin^2(x)cos(x)= cos^3(x)- sin^2(x)cos(x)$.

If x= A/3, then
$sin(3x)= sin(A)= 3sin(A/3)cos^2(A/3)- sin^3(A/3)= 3sin(B)cos^2(B)- sin^3(B)$
and
$cos(3x)= cos(A)= cos^3(A/3)- sin^2(A/3)cos(A/3)= cos^3(B)- sin^2(B)cos(B)$.

From tan A= 12/5, you can solve for sin(A) and cos(A). Then you have two equations to solve for sin(B) and cos(B).