Question

How do I solve this?
tan(theta - 10) = cot(2 theta + 17)

Answer 1

cos(A-B) = cos(A)cos(B) + sin(A)sin(B)
sin(A-B) = sin(A)cos(B) -cos(A)sin(B)
divide them
A=theta
B=10
i hope it will help

Answer 2

^ That maybe effective, there is also a method which goes like this. :)

cot A = tan (90 - A)
\[cot(2\theta+17) = tan(90 - (2\theta+17))\]
And this you know is now \( = tan(\theta - 10)\)

Thus \( = tan(\theta - 10)\) = \(tan(90 - (2\theta+17))\)

NOTE: If tan A = tan B , this implies that, A=B

Now can you find \(\theta\) ? :)

Answer 3

tan (x - 10) = cot (- 2x + 17) = tan (90 - 2x - 17) = tan (-2x + 63)
a) (x - 10) = -2x + 63 -> 3x = 73 -> x = 73/3 degree = 24.20''
b) Add another answer:
tan x = tan (x + 180)
tan (x - 10) = tan (-2x + 63 + 180) --> x - 10 = - 2x + 243
--> 3x = 253 --> x = 253/3 degree