What is the general solution for 1+tanh^2(x)=sech^2(x) ?

The general solution for 1+tanh^2(x)=sech^2(x) is x=0

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1+tanh^2(x)=sech^2(x)

Solution

1 + tanh^{2} (x) = sech^{2} (x)

Solution

x = 0

Degrees

x = 0^{\circ}

Solution steps

1 + tanh^{2} (x) = sech^{2} (x)

Rewrite using trig identities

1 + tanh^{2} (x) = sech^{2} (x)

Use the Hyperbolic identity:

tanh (x) = \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}

1 + (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{2} = sech^{2} (x)

Use the Hyperbolic identity:

sech (x) = \frac{2}{e ^{x} + e ^{- x}}

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

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

1 + (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{2} = (\frac{2}{e ^{x} + e ^{- x}})^{2} : x = 0

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

Apply exponent rules

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

Apply exponent rule:

a^{b c} = (a^{b})^{c}

e^{- x} = (e^{x})^{- 1}

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

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

Rewrite the equation with

e^{x} = u

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

Solve

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

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

Refine

1 + \frac{( u ^{2} - 1 ) ^{2}}{( u ^{2} + 1 ) ^{2}} = \frac{4 u ^{2}}{( u ^{2} + 1 ) ^{2}}

Multiply both sides by

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

1 + \frac{( u ^{2} - 1 ) ^{2}}{( u ^{2} + 1 ) ^{2}} = \frac{4 u ^{2}}{( u ^{2} + 1 ) ^{2}}

Multiply both sides by

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

1 \cdot (u^{2} + 1)^{2} + \frac{( u ^{2} - 1 ) ^{2}}{( u ^{2} + 1 ) ^{2}} (u^{2} + 1)^{2} = \frac{4 u ^{2}}{( u ^{2} + 1 ) ^{2}} (u^{2} + 1)^{2}

Simplify

1 \cdot (u^{2} + 1)^{2} + \frac{( u ^{2} - 1 ) ^{2}}{( u ^{2} + 1 ) ^{2}} (u^{2} + 1)^{2} = \frac{4 u ^{2}}{( u ^{2} + 1 ) ^{2}} (u^{2} + 1)^{2}

Simplify

1 \cdot (u^{2} + 1)^{2} : (u^{2} + 1)^{2}

1 \cdot (u^{2} + 1)^{2}

Multiply:

1 \cdot (u^{2} + 1)^{2} = (u^{2} + 1)^{2}

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

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

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

Simplify

\frac{4 u ^{2}}{( u ^{2} + 1 ) ^{2}} (u^{2} + 1)^{2} : 4 u^{2}

\frac{4 u ^{2}}{( u ^{2} + 1 ) ^{2}} (u^{2} + 1)^{2}

Multiply fractions:

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

= \frac{4 u ^{2} ( u ^{2} + 1 ) ^{2}}{( u ^{2} + 1 ) ^{2}}

Cancel the common factor:

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

= 4 u^{2}

(u^{2} + 1)^{2} + (u^{2} - 1)^{2} = 4 u^{2}

(u^{2} + 1)^{2} + (u^{2} - 1)^{2} = 4 u^{2}

(u^{2} + 1)^{2} + (u^{2} - 1)^{2} = 4 u^{2}

Solve

(u^{2} + 1)^{2} + (u^{2} - 1)^{2} = 4 u^{2} : u = 1, u = - 1

(u^{2} + 1)^{2} + (u^{2} - 1)^{2} = 4 u^{2}

Expand

(u^{2} + 1)^{2} + (u^{2} - 1)^{2} : 2 u^{4} + 2

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

(u^{2} + 1)^{2} : u^{4} + 2 u^{2} + 1

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = u^{2}, b = 1

= (u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2}

Simplify

(u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2} : u^{4} + 2 u^{2} + 1

(u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= (u^{2})^{2} + 2 \cdot 1 \cdot u^{2} + 1

(u^{2})^{2} = u^{4}

(u^{2})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= u^{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= u^{4}

2 u^{2} \cdot 1 = 2 u^{2}

2 u^{2} \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2 u^{2}

= u^{4} + 2 u^{2} + 1

= u^{4} + 2 u^{2} + 1

= u^{4} + 2 u^{2} + 1 + (u^{2} - 1)^{2}

(u^{2} - 1)^{2} : u^{4} - 2 u^{2} + 1

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = u^{2}, b = 1

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

Simplify

(u^{2})^{2} - 2 u^{2} \cdot 1 + 1^{2} : u^{4} - 2 u^{2} + 1

(u^{2})^{2} - 2 u^{2} \cdot 1 + 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

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

(u^{2})^{2} = u^{4}

(u^{2})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= u^{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= u^{4}

2 u^{2} \cdot 1 = 2 u^{2}

2 u^{2} \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2 u^{2}

= u^{4} - 2 u^{2} + 1

= u^{4} - 2 u^{2} + 1

= u^{4} + 2 u^{2} + 1 + u^{4} - 2 u^{2} + 1

Simplify

u^{4} + 2 u^{2} + 1 + u^{4} - 2 u^{2} + 1 : 2 u^{4} + 2

u^{4} + 2 u^{2} + 1 + u^{4} - 2 u^{2} + 1

Group like terms

= u^{4} + u^{4} + 2 u^{2} - 2 u^{2} + 1 + 1

Add similar elements:

2 u^{2} - 2 u^{2} = 0

= u^{4} + u^{4} + 1 + 1

Add similar elements:

u^{4} + u^{4} = 2 u^{4}

= 2 u^{4} + 1 + 1

Add the numbers:

1 + 1 = 2

= 2 u^{4} + 2

= 2 u^{4} + 2

2 u^{4} + 2 = 4 u^{2}

Move

4 u^{2}

to the left side

2 u^{4} + 2 = 4 u^{2}

Subtract

4 u^{2}

from both sides

2 u^{4} + 2 - 4 u^{2} = 4 u^{2} - 4 u^{2}

Simplify

2 u^{4} + 2 - 4 u^{2} = 0

2 u^{4} + 2 - 4 u^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

2 u^{4} - 4 u^{2} + 2 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

2 v^{2} - 4 v + 2 = 0

Solve

2 v^{2} - 4 v + 2 = 0 : v = 1

2 v^{2} - 4 v + 2 = 0

Solve with the quadratic formula

2 v^{2} - 4 v + 2 = 0

Quadratic Equation Formula:

For

a = 2, b = - 4, c = 2

v_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 \cdot 2}{2 \cdot 2}

v_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 \cdot 2}{2 \cdot 2}

(- 4)^{2} - 4 \cdot 2 \cdot 2 = 0

(- 4)^{2} - 4 \cdot 2 \cdot 2

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 4)^{2} = 4^{2}

= 4^{2} - 4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

v_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 2}

v = \frac{- ( - 4 )}{2 \cdot 2}

\frac{- ( - 4 )}{2 \cdot 2} = 1

\frac{- ( - 4 )}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{4}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

v = 1

The solution to the quadratic equation is:

v = 1

v = 1

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

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

u = 1, u = - 1

1 = 1

1

Apply radical rule:

1 = 1

= 1

- 1 = - 1

- 1

Apply radical rule:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

The solutions are

u = 1, u = - 1

u = 1, u = - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

and compare to zero

u = 0

Take the denominator(s) of

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

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 1, u = - 1

u = 1, u = - 1

Substitute back

u = e^{x},

solve for

x

Solve

e^{x} = 1 : x = 0

e^{x} = 1

Apply exponent rules

e^{x} = 1

f (x) = g (x)

, then

ln (f (x)) = ln (g (x))

ln (e^{x}) = ln (1)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (1)

Simplify

ln (1) : 0

ln (1)

Apply log rule:

lo g_{a} (1) = 0

= 0

x = 0

x = 0

Solve

e^{x} = - 1 :

No Solution for

x \in R

e^{x} = - 1

a^{f (x)}

cannot be zero or negative for

x \in R

No Solution for x \in R

x = 0

x = 0

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Frequently Asked Questions (FAQ)

What is the general solution for 1+tanh^2(x)=sech^2(x) ?
The general solution for 1+tanh^2(x)=sech^2(x) is x=0