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

The general solution for (tan^2(b)+1)/(tan(b))=csc^2(b) is b= pi/4+pin

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

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

b = \frac{π}{4} + πn

Solution steps

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

Subtract

csc^{2} (b)

from both sides

\frac{tan ^{2} ( b ) + 1}{tan ( b )} - csc^{2} (b) = 0

Simplify

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

\frac{tan ^{2} ( b ) + 1 - csc ^{2} ( b ) tan ( b )}{tan ( b )} = 0

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

tan^{2} (b) + 1 - csc^{2} (b) tan (b) = 0

Rewrite using trig identities

1 - \frac{1 + cot ^{2} ( b )}{cot ( b )} + \frac{1}{cot ^{2} ( b )} = 0

Solve by substitution

cot (b) = 1, cot (b) = i, cot (b) = - i

cot (b) = 1 : b = \frac{π}{4} + πn

cot (b) = i :

No Solution

cot (b) = - i :

No Solution

Combine all the solutions

b = \frac{π}{4} + πn

x, r + s + 6 t = cos (2 x + y)

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

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

What is the general solution for (tan^2(b)+1)/(tan(b))=csc^2(b) ?
The general solution for (tan^2(b)+1)/(tan(b))=csc^2(b) is b= pi/4+pin