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

The general solution for 1+sec^2(x)=tan^2(x) is No Solution for x\in\mathbb{R}

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Popular >

1+sec^2(x)=tan^2(x)

Solution

1 + sec^{2} (x) = tan^{2} (x)

Solution

No Solution for x \in R

Solution steps

1 + sec^{2} (x) = tan^{2} (x)

Subtract

tan^{2} (x)

from both sides

1 + sec^{2} (x) - tan^{2} (x) = 0

Express with sin, cos

1 + sec^{2} (x) - tan^{2} (x)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

Simplify

1 + (\frac{1}{cos ( x )})^{2} - (\frac{sin ( x )}{cos ( x )})^{2} : \frac{cos ^{2} ( x ) + 1 - sin ^{2} ( x )}{cos ^{2} ( x )}

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

(\frac{1}{cos ( x )})^{2} = \frac{1}{cos ^{2} ( x )}

(\frac{1}{cos ( x )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( x )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( x )}

(\frac{sin ( x )}{cos ( x )})^{2} = \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

(\frac{sin ( x )}{cos ( x )})^{2}

Apply exponent rule:

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

= \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

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

Combine the fractions

\frac{1}{cos ^{2} ( x )} - \frac{sin ^{2} ( x )}{cos ^{2} ( x )} : \frac{1 - sin ^{2} ( x )}{cos ^{2} ( x )}

Apply rule

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

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

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

Convert element to fraction:

1 = \frac{1 cos ^{2} ( x )}{cos ^{2} ( x )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

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

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

\frac{1 + cos ^{2} ( x ) - sin ^{2} ( x )}{cos ^{2} ( x )} = 0

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

1 + cos^{2} (x) - sin^{2} (x) = 0

Rewrite using trig identities

1 + cos^{2} (x) - sin^{2} (x)

Use the Double Angle identity:

cos^{2} (x) - sin^{2} (x) = cos (2 x)

= 1 + cos (2 x)

1 + cos (2 x) = 0

Move

1

to the right side

1 + cos (2 x) = 0

Subtract

1

from both sides

1 + cos (2 x) - 1 = 0 - 1

Simplify

cos (2 x) = - 1

cos (2 x) = - 1

General solutions for

cos (2 x) = - 1

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 x = π + 2 πn

2 x = π + 2 πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

Since the equation is undefined for:

\frac{π}{2} + πn

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for 1+sec^2(x)=tan^2(x) ?
The general solution for 1+sec^2(x)=tan^2(x) is No Solution for x\in\mathbb{R}