Is sec(pi/2-y)=csc(y) ?

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

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prove sec(pi/2-y)=csc(y)

Solution

prove

sec (\frac{π}{2} - y) = csc (y)

Solution

True

Solution steps

sec (\frac{π}{2} - y) = csc (y)

Manipulating left side

sec (\frac{π}{2} - y)

Rewrite using trig identities

sec (\frac{π}{2} - y)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( \frac{π}{2} - y )}

Use the Angle Difference identity:

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

= \frac{1}{cos ( \frac{π}{2} ) cos ( y ) + sin ( \frac{π}{2} ) sin ( y )}

Simplify

\frac{1}{cos ( \frac{π}{2} ) cos ( y ) + sin ( \frac{π}{2} ) sin ( y )} : \frac{1}{sin ( y )}

\frac{1}{cos ( \frac{π}{2} ) cos ( y ) + sin ( \frac{π}{2} ) sin ( y )}

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

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

cos (\frac{π}{2}) cos (y) = 0

cos (\frac{π}{2}) cos (y)

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 (y)

Apply rule

0 \cdot a = 0

= 0

sin (\frac{π}{2}) sin (y) = sin (y)

sin (\frac{π}{2}) sin (y)

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 (y)

Multiply:

1 \cdot sin (y) = sin (y)

= sin (y)

= 0 + sin (y)

0 + sin (y) = sin (y)

= sin (y)

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= csc (y)

csc (y)

csc (y)

We showed that the two sides could take the same form

\Rightarrow True

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