What is the general solution for (cot(B))/(cos(B))+1/(sin(B))=1 ?

The general solution for (cot(B))/(cos(B))+1/(sin(B))=1 is No Solution for B\in\mathbb{R}

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(cot(B))/(cos(B))+1/(sin(B))=1

\frac{cot ( B )}{cos ( B )} + \frac{1}{sin ( B )} = 1

No Solution for B \in R

Solution steps

\frac{cot ( B )}{cos ( B )} + \frac{1}{sin ( B )} = 1

Subtract

1

from both sides

\frac{cot ( B )}{cos ( B )} + \frac{1}{sin ( B )} - 1 = 0

Simplify

\frac{cot ( B )}{cos ( B )} + \frac{1}{sin ( B )} - 1 : \frac{cot ( B ) sin ( B ) + cos ( B ) - cos ( B ) sin ( B )}{cos ( B ) sin ( B )}

\frac{cot ( B ) sin ( B ) + cos ( B ) - cos ( B ) sin ( B )}{cos ( B ) sin ( B )} = 0

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

cot (B) sin (B) + cos (B) - cos (B) sin (B) = 0

Rewrite using trig identities

2 cos (B) - cos (B) sin (B) = 0

Factor

2 cos (B) - cos (B) sin (B) : - cos (B) (sin (B) - 2)

- cos (B) (sin (B) - 2) = 0

Solving each part separately

cos (B) = 0 or sin (B) - 2 = 0

cos (B) = 0 : B = \frac{π}{2} + 2 πn, B = \frac{3 π}{2} + 2 πn

sin (B) - 2 = 0 :

No Solution

Combine all the solutions

B = \frac{π}{2} + 2 πn, B = \frac{3 π}{2} + 2 πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn

No Solution for B \in R

cos (3 x - \frac{π}{3}) = 1

2 sin (2 x - 10) - 3 = 0

2 sin (x) + (2 + 2) = - 2 csc (x)

\frac{1 - cos ( x )}{2} + cos (x) - 1 = 0

arctan (x) = \frac{π}{6}

What is the general solution for (cot(B))/(cos(B))+1/(sin(B))=1 ?
The general solution for (cot(B))/(cos(B))+1/(sin(B))=1 is No Solution for B\in\mathbb{R}