What is the general solution for tan(x)=sec(x) ?

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

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tan(x)=sec(x)

tan (x) = sec (x)

No Solution for x \in R

Solution steps

tan (x) = sec (x)

Subtract

sec (x)

from both sides

tan (x) - sec (x) = 0

Express with sin, cos

- sec (x) + tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply rule

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

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

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

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

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

- 1 + sin (x) = 0

Move

1

to the right side

- 1 + sin (x) = 0

Add

1

to both sides

- 1 + sin (x) + 1 = 0 + 1

Simplify

sin (x) = 1

sin (x) = 1

General solutions for

sin (x) = 1

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Since the equation is undefined for:

\frac{π}{2} + 2 πn

No Solution for x \in R

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tan (4 5^{\circ} - x) + tan (x) = 1

tan (3 B + 5^{\circ}) = cot (2 B + 1 0^{\circ})

2 cot (x) cos (x) + 7 = 7 csc (x)

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