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

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

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Popular >

tan(x)cot(x)=csc(x)sec(x)

Solution

tan (x) cot (x) = csc (x) sec (x)

Solution

No Solution for x \in R

Solution steps

tan (x) cot (x) = csc (x) sec (x)

Subtract

csc (x) sec (x)

from both sides

tan (x) cot (x) - csc (x) sec (x) = 0

Express with sin, cos

cot (x) tan (x) - csc (x) sec (x)

Use the basic trigonometric identity:

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

= \frac{cos ( x )}{sin ( x )} tan (x) - csc (x) sec (x)

Use the basic trigonometric identity:

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

= \frac{cos ( x )}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )} - csc (x) sec (x)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

\frac{cos ( x )}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )} = 1

\frac{cos ( x )}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )}

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

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

Cancel the common factor:

sin (x)

= 1

\frac{1}{sin ( x )} \cdot \frac{1}{cos ( x )} = \frac{1}{sin ( x ) cos ( x )}

\frac{1}{sin ( x )} \cdot \frac{1}{cos ( x )}

Multiply fractions:

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

= \frac{1 \cdot 1}{sin ( x ) cos ( x )}

Multiply the numbers:

1 \cdot 1 = 1

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

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

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

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

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

Rewrite using trig identities

- 1 + cos (x) sin (x)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

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

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

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

Move

1

to the right side

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

Add

1

to both sides

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

Simplify

\frac{sin ( 2 x )}{2} = 1

\frac{sin ( 2 x )}{2} = 1

Multiply both sides by

2

\frac{sin ( 2 x )}{2} = 1

Multiply both sides by

2

\frac{2 sin ( 2 x )}{2} = 1 \cdot 2

Simplify

sin (2 x) = 2

sin (2 x) = 2

- 1 \leq sin (x) \leq 1

No Solution for x \in R

Graph

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cos (A) = \frac{3}{5}

cos^2(x)-sin^2(x)= 1/(sqrt(2))

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

arctan(2x)+arctan(x)= pi/2

arctan (2 x) + arctan (x) = \frac{π}{2}

sec(b/4)csc(b/4)=2csc(b/4)

sec (\frac{b}{4}) csc (\frac{b}{4}) = 2 csc (\frac{b}{4})

2cos(x)-5=3cos(x)-4

2 cos (x) - 5 = 3 cos (x) - 4

Frequently Asked Questions (FAQ)

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