What is the general solution for (tan^2(b)+1)/((tan(x)))=csc^2(b) ?

The general solution for (tan^2(b)+1)/((tan(x)))=csc^2(b) is x=arctan((tan^2(b)+1)/(csc^2(b)))+pin

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(tan^2(b)+1)/((tan(x)))=csc^2(b)

\frac{tan ^{2} ( b ) + 1}{( tan ( x ) )} = csc^{2} (b)

x = arctan (\frac{tan ^{2} ( b ) + 1}{csc ^{2} ( b )}) + πn

Solution steps

\frac{tan ^{2} ( b ) + 1}{( tan ( x ) )} = csc^{2} (b)

Multiply both sides by

tan (x)

\frac{tan ^{2} ( b ) + 1}{tan ( x )} = csc^{2} (b)

Multiply both sides by

tan (x)

\frac{tan ^{2} ( b ) + 1}{tan ( x )} tan (x) = csc^{2} (b) tan (x)

Simplify

tan^{2} (b) + 1 = csc^{2} (b) tan (x)

tan^{2} (b) + 1 = csc^{2} (b) tan (x)

Switch sides

csc^{2} (b) tan (x) = tan^{2} (b) + 1

Divide both sides by

csc^{2} (b)

csc^{2} (b) tan (x) = tan^{2} (b) + 1

Divide both sides by

csc^{2} (b)

\frac{csc ^{2} ( b ) tan ( x )}{csc ^{2} ( b )} = \frac{tan ^{2} ( b )}{csc ^{2} ( b )} + \frac{1}{csc ^{2} ( b )}

Simplify

\frac{csc ^{2} ( b ) tan ( x )}{csc ^{2} ( b )} = \frac{tan ^{2} ( b )}{csc ^{2} ( b )} + \frac{1}{csc ^{2} ( b )}

Simplify

\frac{csc ^{2} ( b ) tan ( x )}{csc ^{2} ( b )} : tan (x)

\frac{csc ^{2} ( b ) tan ( x )}{csc ^{2} ( b )}

Cancel the common factor:

csc^{2} (b)

= tan (x)

Simplify

\frac{tan ^{2} ( b )}{csc ^{2} ( b )} + \frac{1}{csc ^{2} ( b )} : \frac{tan ^{2} ( b ) + 1}{csc ^{2} ( b )}

\frac{tan ^{2} ( b )}{csc ^{2} ( b )} + \frac{1}{csc ^{2} ( b )}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{tan ^{2} ( b ) + 1}{csc ^{2} ( b )}

tan (x) = \frac{tan ^{2} ( b ) + 1}{csc ^{2} ( b )}

tan (x) = \frac{tan ^{2} ( b ) + 1}{csc ^{2} ( b )}

tan (x) = \frac{tan ^{2} ( b ) + 1}{csc ^{2} ( b )}

Apply trig inverse properties

tan (x) = \frac{tan ^{2} ( b ) + 1}{csc ^{2} ( b )}

General solutions for

tan (x) = \frac{tan ^{2} ( b ) + 1}{csc ^{2} ( b )}

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

x = arctan (\frac{tan ^{2} ( b ) + 1}{csc ^{2} ( b )}) + πn

x = arctan (\frac{tan ^{2} ( b ) + 1}{csc ^{2} ( b )}) + πn

sin (x) + sin^{2} (\frac{x}{2}) = \frac{1}{2}

sin^{5} (x) + sin^{3} (x) = 0

5 sin^{2} (x) cos (7 x) - cos (7 x) = 0

x, cos^{2} (x) \cdot z^{2} - 2 cos^{2} (x) \cdot z + 1 = 0

sin (a) + sin (12 0^{\circ} + a) + sin (12 0^{\circ} - a) = 0

What is the general solution for (tan^2(b)+1)/((tan(x)))=csc^2(b) ?
The general solution for (tan^2(b)+1)/((tan(x)))=csc^2(b) is x=arctan((tan^2(b)+1)/(csc^2(b)))+pin