Is tan(pi/2-θ)tan(θ)=1 ?

The answer to whether tan(pi/2-θ)tan(θ)=1 is True

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Popular Trigonometry >

prove tan(pi/2-θ)tan(θ)=1

Solution

prove

tan (\frac{π}{2} - θ) tan (θ) = 1

Solution

True

Solution steps

tan (\frac{π}{2} - θ) tan (θ) = 1

Manipulating left side

tan (\frac{π}{2} - θ) tan (θ)

Rewrite using trig identities

tan (\frac{π}{2} - θ)

Use the basic trigonometric identity:

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

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

Use the Angle Difference identity:

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

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

Use the Angle Difference identity:

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

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

Simplify

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

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

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

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

sin (\frac{π}{2}) cos (θ) = cos (θ)

sin (\frac{π}{2}) cos (θ)

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 (θ)

Multiply:

1 \cdot cos (θ) = cos (θ)

= cos (θ)

cos (\frac{π}{2}) sin (θ) = 0

cos (\frac{π}{2}) sin (θ)

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 (θ)

Apply rule

0 \cdot a = 0

= 0

= cos (θ) - 0

cos (θ) - 0 = cos (θ)

= cos (θ)

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

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

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

cos (\frac{π}{2}) cos (θ) = 0

cos (\frac{π}{2}) cos (θ)

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 (θ)

Apply rule

0 \cdot a = 0

= 0

sin (\frac{π}{2}) sin (θ) = sin (θ)

sin (\frac{π}{2}) sin (θ)

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 (θ)

Multiply:

1 \cdot sin (θ) = sin (θ)

= sin (θ)

= 0 + sin (θ)

0 + sin (θ) = sin (θ)

= sin (θ)

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )} tan (θ)

Multiply fractions:

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

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

Express with sin, cos

\frac{cos ( θ ) tan ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

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

\frac{cos ( θ ) \frac{s i n ( θ )}{c o s ( θ )}}{sin ( θ )} = 1

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

Multiply

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

cos (θ) \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= sin (θ)

= \frac{sin ( θ )}{sin ( θ )}

Apply rule

\frac{a}{a} = 1

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is tan(pi/2-θ)tan(θ)=1 ?
The answer to whether tan(pi/2-θ)tan(θ)=1 is True