Is tan(pi/2-x)sec(x)=csc(x) ?

The answer to whether tan(pi/2-x)sec(x)=csc(x) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove tan(pi/2-x)sec(x)=csc(x)

Solution

prove

tan (\frac{π}{2} - x) sec (x) = csc (x)

Solution

True

Solution steps

tan (\frac{π}{2} - x) sec (x) = csc (x)

Manipulating left side

tan (\frac{π}{2} - x) sec (x)

Rewrite using trig identities

tan (\frac{π}{2} - x)

Use the basic trigonometric identity:

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

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

Use the Angle Difference identity:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

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

Use the Angle Difference identity:

cos (s - t) = cos (s) cos (t) + sin (s) sin (t)

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

Simplify

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

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

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

sin (\frac{π}{2}) cos (x) - cos (\frac{π}{2}) sin (x)

sin (\frac{π}{2}) cos (x) = cos (x)

sin (\frac{π}{2}) cos (x)

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 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}

= 1

= 1 \cdot cos (x)

Multiply:

1 \cdot cos (x) = cos (x)

= cos (x)

cos (\frac{π}{2}) sin (x) = 0

cos (\frac{π}{2}) sin (x)

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

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}

= 0

= 0 \cdot sin (x)

Apply rule

0 \cdot a = 0

= 0

= cos (x) - 0

cos (x) - 0 = cos (x)

= cos (x)

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

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

cos (\frac{π}{2}) cos (x) + sin (\frac{π}{2}) sin (x)

cos (\frac{π}{2}) cos (x) = 0

cos (\frac{π}{2}) cos (x)

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

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}

= 0

= 0 \cdot cos (x)

Apply rule

0 \cdot a = 0

= 0

sin (\frac{π}{2}) sin (x) = sin (x)

sin (\frac{π}{2}) sin (x)

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 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}

= 1

= 1 \cdot sin (x)

Multiply:

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

= sin (x)

= 0 + sin (x)

0 + sin (x) = sin (x)

= sin (x)

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

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

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

Multiply fractions:

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

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{cos ( x ) \frac{1}{c o s ( x )}}{sin ( x )}

Simplify

\frac{cos ( x ) \frac{1}{c o s ( x )}}{sin ( x )} : \frac{1}{sin ( x )}

\frac{cos ( x ) \frac{1}{c o s ( x )}}{sin ( x )}

Multiply

cos (x) \frac{1}{cos ( x )} : 1

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= 1

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( x )}}

Simplify

\frac{1}{\frac{1}{c s c ( x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

Apply rule

\frac{a}{1} = a

= csc (x)

csc (x)

csc (x)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove 1/(cos(x))-cos(x)=sin(x)tan(x)prove

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

prove sec(u)cos(u)-sin^2(u)=cos^2(u)prove

sec (u) cos (u) - sin^{2} (u) = cos^{2} (u)

prove 1+cot^2(x/2)= 2/(sin(x)tan(x/2))prove

1 + cot^{2} (\frac{x}{2}) = \frac{2}{sin ( x ) tan ( \frac{x}{2} )}

prove (tan^2(x))/(sec(x))+cos(x)=sec(x)prove

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

prove tan^2(x)cot^2(x)=1 prove

tan^{2} (x) cot^{2} (x) = 1

Frequently Asked Questions (FAQ)

Is tan(pi/2-x)sec(x)=csc(x) ?
The answer to whether tan(pi/2-x)sec(x)=csc(x) is True