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Popular >

tan(x)<tan(y)

Solution

tan (x) < tan (y)

False for all x \in R

Solution steps

tan (x) < tan (y)

Move

tan (y)

to the left side

tan (x) < tan (y)

Subtract

tan (y)

from both sides

tan (x) - tan (y) < tan (y) - tan (y)

tan (x) - tan (y) < 0

tan (x) - tan (y) < 0

Periodicity of

tan (x) - tan (y) :

Not periodic

The function

tan (x) - tan (y)

is not periodic

= Not periodic

Express with sin, cos

tan (x) - tan (y) < 0

Use the basic trigonometric identity:

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

\frac{sin ( x )}{cos ( x )} - tan (y) < 0

Use the basic trigonometric identity:

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

\frac{sin ( x )}{cos ( x )} - \frac{sin ( y )}{cos ( y )} < 0

\frac{sin ( x )}{cos ( x )} - \frac{sin ( y )}{cos ( y )} < 0

Simplify

\frac{sin ( x )}{cos ( x )} - \frac{sin ( y )}{cos ( y )} : \frac{cos ( y ) sin ( x ) - sin ( y ) cos ( x )}{cos ( y ) cos ( x )}

\frac{sin ( x )}{cos ( x )} - \frac{sin ( y )}{cos ( y )}

Least Common Multiplier of

cos (x), cos (y) : cos (y) cos (x)

cos (x), cos (y)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (x)

cos (y)

= cos (y) cos (x)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

cos (y) cos (x)

For

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

multiply the denominator and numerator by

cos (y)

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

For

\frac{sin ( y )}{cos ( y )} :

multiply the denominator and numerator by

cos (x)

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

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

Since the denominators are equal, combine the fractions:

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

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

= No Solution for x \in R

False for all x \in R

tan (x - \frac{π}{6}) \geq - 1

sin (\frac{x}{3}) < \frac{3}{2}

tan (x) \geq - \frac{3}{3}

(2 sin (2 x) + \frac{1}{2}) \leq \frac{1}{2}

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