What is the general solution for 1+cot^2(a)=tan^2(a) ?

The general solution for 1+cot^2(a)=tan^2(a) is a=0.90455…+pin,a=2.23703…+pin

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1+cot^2(a)=tan^2(a)

1 + cot^{2} (a) = tan^{2} (a)

a = 0.90455 \dots + πn, a = 2.23703 \dots + πn

Solution steps

1 + cot^{2} (a) = tan^{2} (a)

Subtract

tan^{2} (a)

from both sides

1 + cot^{2} (a) - tan^{2} (a) = 0

Rewrite using trig identities

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

Solve by substitution

cot (a) = \frac{- 1 + 5}{2}, cot (a) = - \frac{- 1 + 5}{2}, cot (a) = \frac{- 1 - 5}{2}, cot (a) = - \frac{- 1 - 5}{2}

cot (a) = \frac{- 1 + 5}{2} : a = arccot \frac{- 1 + 5}{2} + πn

cot (a) = - \frac{- 1 + 5}{2} : a = arccot - \frac{- 1 + 5}{2} + πn

cot (a) = \frac{- 1 - 5}{2} : a = arccot \frac{- 1 - 5}{2} + πn

cot (a) = - \frac{- 1 - 5}{2} : a = arccot - \frac{- 1 - 5}{2} + πn

Combine all the solutions

a = arccot \frac{- 1 + 5}{2} + πn, a = arccot - \frac{- 1 + 5}{2} + πn, a = arccot \frac{- 1 - 5}{2} + πn, a = arccot - \frac{- 1 - 5}{2} + πn

Since the equation is undefined for:

arccot \frac{- 1 - 5}{2} + πn, arccot - \frac{- 1 - 5}{2} + πn

a = arccot \frac{- 1 + 5}{2} + πn, a = arccot - \frac{- 1 + 5}{2} + πn

Show solutions in decimal form

a = 0.90455 \dots + πn, a = 2.23703 \dots + πn

tan (α) = \frac{12}{6 3}

1 = cos^{2} (α) - 2 cos (α) sin (a) + sin^{2} (α)

4 sin (θ) = 1

sin (x + \frac{π}{2}) = 0.6

4 cot (x) + 1 = 1 + 2 cot (x)

What is the general solution for 1+cot^2(a)=tan^2(a) ?
The general solution for 1+cot^2(a)=tan^2(a) is a=0.90455…+pin,a=2.23703…+pin