What is the general solution for tan(b+15)*cot(2b-5)=1 ?

The general solution for tan(b+15)*cot(2b-5)=1 is b=-2pin+20,b=-pi+20-2pin

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tan(b+15)*cot(2b-5)=1

tan (b + 15) \cdot cot (2 b - 5) = 1

b = - 2 πn + 20, b = - π + 20 - 2 πn

Solution steps

tan (b + 15) cot (2 b - 5) = 1

Subtract

1

from both sides

tan (b + 15) cot (2 b - 5) - 1 = 0

Express with sin, cos

\frac{cos ( - 5 + 2 b ) sin ( 15 + b ) - cos ( 15 + b ) sin ( - 5 + 2 b )}{cos ( 15 + b ) sin ( - 5 + 2 b )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

cos (- 5 + 2 b) sin (15 + b) - cos (15 + b) sin (- 5 + 2 b) = 0

Rewrite using trig identities

sin (15 + b - (- 5 + 2 b)) = 0

General solutions for

sin (15 + b - (- 5 + 2 b)) = 0

15 + b - (- 5 + 2 b) = 0 + 2 πn, 15 + b - (- 5 + 2 b) = π + 2 πn

Solve

15 + b - (- 5 + 2 b) = 0 + 2 πn : b = - 2 πn + 20

Solve

15 + b - (- 5 + 2 b) = π + 2 πn : b = - π + 20 - 2 πn

b = - 2 πn + 20, b = - π + 20 - 2 πn

cos (x) = \frac{5}{3}

cos (x) = - \frac{7}{8}

tan (x) = - \frac{3}{5}

sin (x) = - cos (2 x)

csc (\frac{π}{24} x) = 1

What is the general solution for tan(b+15)*cot(2b-5)=1 ?
The general solution for tan(b+15)*cot(2b-5)=1 is b=-2pin+20,b=-pi+20-2pin