Is sin(t)csc(pi/2-t)=tan(t) ?

The answer to whether sin(t)csc(pi/2-t)=tan(t) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

prove sin(t)csc(pi/2-t)=tan(t)

prove

sin (t) csc (\frac{π}{2} - t) = tan (t)

True

Solution steps

sin (t) csc (\frac{π}{2} - t) = tan (t)

Manipulating left side

sin (t) csc (\frac{π}{2} - t)

Rewrite using trig identities

csc (\frac{π}{2} - t)

Use the basic trigonometric identity:

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

= \frac{1}{sin ( \frac{π}{2} - t )}

Use the Angle Difference identity:

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

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

Simplify

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

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

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

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

sin (\frac{π}{2}) cos (t) = cos (t)

sin (\frac{π}{2}) cos (t)

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 (t)

Multiply:

1 \cdot cos (t) = cos (t)

= cos (t)

cos (\frac{π}{2}) sin (t) = 0

cos (\frac{π}{2}) sin (t)

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 (t)

Apply rule

0 \cdot a = 0

= 0

= cos (t) - 0

cos (t) - 0 = cos (t)

= cos (t)

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

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

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

Simplify

sin (t) \frac{1}{cos ( t )} : \frac{sin ( t )}{cos ( t )}

sin (t) \frac{1}{cos ( t )}

Multiply fractions:

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

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

Multiply:

1 \cdot sin (t) = sin (t)

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

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

Use the basic trigonometric identity:

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

= tan (t)

We showed that the two sides could take the same form

\Rightarrow True

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