What is the general solution for a=((1+sin^2(x)))/((1-sin^2(x))) ?

The general solution for a=((1+sin^2(x)))/((1-sin^2(x))) is

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Popular Trigonometry >

a=((1+sin^2(x)))/((1-sin^2(x)))

Solution

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

Solution

x = arcsin (\frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (\frac{- 1 + a}{1 + a}) + 2 πn

Solution steps

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

Switch sides

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

Solve by substitution

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

Let:

sin (x) = u

\frac{1 + u ^{2}}{1 - u ^{2}} = a

\frac{1 + u ^{2}}{1 - u ^{2}} = a : u = \frac{- 1 + a}{1 + a}, u = - \frac{- 1 + a}{1 + a}; a \neq = - 1

\frac{1 + u ^{2}}{1 - u ^{2}} = a

Multiply both sides by

1 - u^{2}

\frac{1 + u ^{2}}{1 - u ^{2}} = a

Multiply both sides by

1 - u^{2}

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

Simplify

1 + u^{2} = a (1 - u^{2})

1 + u^{2} = a (1 - u^{2})

Solve

1 + u^{2} = a (1 - u^{2}) : u = \frac{- 1 + a}{1 + a}, u = - \frac{- 1 + a}{1 + a}; a \neq = - 1

1 + u^{2} = a (1 - u^{2})

Move

1

to the right side

1 + u^{2} = a (1 - u^{2})

Subtract

1

from both sides

1 + u^{2} - 1 = a (1 - u^{2}) - 1

Simplify

u^{2} = a (1 - u^{2}) - 1

u^{2} = a (1 - u^{2}) - 1

Move

a (1 - u^{2})

to the left side

u^{2} = a (1 - u^{2}) - 1

Subtract

a (1 - u^{2})

from both sides

u^{2} - a (1 - u^{2}) = a (1 - u^{2}) - 1 - a (1 - u^{2})

Simplify

u^{2} - a (1 - u^{2}) = - 1

u^{2} - a (1 - u^{2}) = - 1

Expand

- a (1 - u^{2}) : - a + a u^{2}

- a (1 - u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = - a, b = 1, c = u^{2}

= - a \cdot 1 - (- a) u^{2}

Apply minus-plus rules

- (- a) = a

= - 1 \cdot a + a u^{2}

Multiply:

1 \cdot a = a

= - a + a u^{2}

u^{2} - a + a u^{2} = - 1

Move

a

to the right side

u^{2} - a + a u^{2} = - 1

Add

a

to both sides

u^{2} - a + a u^{2} + a = - 1 + a

Simplify

u^{2} + a u^{2} = - 1 + a

u^{2} + a u^{2} = - 1 + a

Factor

u^{2} + a u^{2} : u^{2} (1 + a)

u^{2} + a u^{2}

Factor out common term

u^{2}

= u^{2} (1 + a)

u^{2} (1 + a) = - 1 + a

Divide both sides by

1 + a; a \neq = - 1

u^{2} (1 + a) = - 1 + a

Divide both sides by

1 + a; a \neq = - 1

\frac{u ^{2} ( 1 + a )}{1 + a} = - \frac{1}{1 + a} + \frac{a}{1 + a}; a \neq = - 1

Simplify

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

Simplify

\frac{u ^{2} ( 1 + a )}{1 + a} : u^{2}

\frac{u ^{2} ( 1 + a )}{1 + a}

Cancel the common factor:

1 + a

= u^{2}

Simplify

- \frac{1}{1 + a} + \frac{a}{1 + a} : \frac{- 1 + a}{1 + a}

- \frac{1}{1 + a} + \frac{a}{1 + a}

Apply rule

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

= \frac{- 1 + a}{1 + a}

u^{2} = \frac{- 1 + a}{1 + a}; a \neq = - 1

u^{2} = \frac{- 1 + a}{1 + a}; a \neq = - 1

u^{2} = \frac{- 1 + a}{1 + a}; a \neq = - 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{- 1 + a}{1 + a}, u = - \frac{- 1 + a}{1 + a}; a \neq = - 1

u = \frac{- 1 + a}{1 + a}, u = - \frac{- 1 + a}{1 + a}; a \neq = - 1

Substitute back

u = sin (x)

sin (x) = \frac{- 1 + a}{1 + a}, sin (x) = - \frac{- 1 + a}{1 + a}; a \neq = - 1

sin (x) = \frac{- 1 + a}{1 + a}, sin (x) = - \frac{- 1 + a}{1 + a}; a \neq = - 1

sin (x) = \frac{- 1 + a}{1 + a} : x = arcsin (\frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn

sin (x) = \frac{- 1 + a}{1 + a}

Apply trig inverse properties

sin (x) = \frac{- 1 + a}{1 + a}

General solutions for

sin (x) = \frac{- 1 + a}{1 + a}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (\frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn

x = arcsin (\frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn

sin (x) = - \frac{- 1 + a}{1 + a} : x = arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (\frac{- 1 + a}{1 + a}) + 2 πn

sin (x) = - \frac{- 1 + a}{1 + a}

Apply trig inverse properties

sin (x) = - \frac{- 1 + a}{1 + a}

General solutions for

sin (x) = - \frac{- 1 + a}{1 + a}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (\frac{- 1 + a}{1 + a}) + 2 πn

x = arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (\frac{- 1 + a}{1 + a}) + 2 πn

Combine all the solutions

x = arcsin (\frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (\frac{- 1 + a}{1 + a}) + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for a=((1+sin^2(x)))/((1-sin^2(x))) ?
The general solution for a=((1+sin^2(x)))/((1-sin^2(x))) is