What is the general solution for solvefor A,(1+tan(A))(1+tan(B))=2 ?

The general solution for solvefor A,(1+tan(A))(1+tan(B))=2 is A=arctan(2/(1+tan(B))-1)+pin

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solvefor A,(1+tan(A))(1+tan(B))=2

solve for

A, (1 + tan (A)) (1 + tan (B)) = 2

A = arctan (\frac{2}{1 + tan ( B )} - 1) + πn

Solution steps

(1 + tan (A)) (1 + tan (B)) = 2

Divide both sides by

1 + tan (B); B \neq = \frac{3 π}{4} + πn

(1 + tan (A)) (1 + tan (B)) = 2

Divide both sides by

1 + tan (B); B \neq = \frac{3 π}{4} + πn

\frac{( 1 + tan ( A ) ) ( 1 + tan ( B ) )}{1 + tan ( B )} = \frac{2}{1 + tan ( B )}; B \neq = \frac{3 π}{4} + πn

Simplify

1 + tan (A) = \frac{2}{1 + tan ( B )}; B \neq = \frac{3 π}{4} + πn

1 + tan (A) = \frac{2}{1 + tan ( B )}; B \neq = \frac{3 π}{4} + πn

Move

1

to the right side

1 + tan (A) = \frac{2}{1 + tan ( B )}; B \neq = \frac{3 π}{4} + πn

Subtract

1

from both sides

1 + tan (A) - 1 = \frac{2}{1 + tan ( B )} - 1; B \neq = \frac{3 π}{4} + πn

Simplify

tan (A) = \frac{2}{1 + tan ( B )} - 1; B \neq = \frac{3 π}{4} + πn

tan (A) = \frac{2}{1 + tan ( B )} - 1; B \neq = \frac{3 π}{4} + πn

Apply trig inverse properties

tan (A) = \frac{2}{1 + tan ( B )} - 1

General solutions for

tan (A) = \frac{2}{1 + tan ( B )} - 1

tan (x) = a \Rightarrow x = arctan (a) + πn

A = arctan (\frac{2}{1 + tan ( B )} - 1) + πn

A = arctan (\frac{2}{1 + tan ( B )} - 1) + πn

5 cos (2 t) cos (t) = 0

k, 6 (- cos (\frac{k}{2}) + 1) = 1.5

tan (x) sec (x) = 2 tan (x)

tan (x) = \frac{5.1}{10.5}

tan (x) = 3, 0 \leq x < 2 π

What is the general solution for solvefor A,(1+tan(A))(1+tan(B))=2 ?
The general solution for solvefor A,(1+tan(A))(1+tan(B))=2 is A=arctan(2/(1+tan(B))-1)+pin