Is csc(pi/2-x)cot(x)=csc(x) ?

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

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Popular Trigonometry >

prove csc(pi/2-x)cot(x)=csc(x)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

csc (\frac{π}{2} - x) cot (x)

Rewrite using trig identities

csc (\frac{π}{2} - x)

Use the basic trigonometric identity:

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

= \frac{1}{sin ( \frac{π}{2} - x )}

Use the Angle Difference identity:

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

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

Simplify

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

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

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

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

Simplify

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot cot (x) = cot (x)

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

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{\frac{c o s ( x )}{s i n ( x )}}{cos ( x )}

Simplify

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

\frac{\frac{c o s ( x )}{s i n ( x )}}{cos ( x )}

Apply the fraction rule:

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

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

Cancel the common factor:

cos (x)

= \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

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Frequently Asked Questions (FAQ)

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