Is sec^2(y)-cot^2(pi/2-y)=1 ?

The answer to whether sec^2(y)-cot^2(pi/2-y)=1 is True

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Popular Trigonometry >

prove sec^2(y)-cot^2(pi/2-y)=1

Solution

prove

sec^{2} (y) - cot^{2} (\frac{π}{2} - y) = 1

Solution

True

Solution steps

sec^{2} (y) - cot^{2} (\frac{π}{2} - y) = 1

Manipulating left side

sec^{2} (y) - cot^{2} (\frac{π}{2} - y)

Rewrite using trig identities

cot (\frac{π}{2} - y)

Use the basic trigonometric identity:

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

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

Use the Angle Difference identity:

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

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

Use the Angle Difference identity:

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

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

Simplify

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

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

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

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

sin (\frac{π}{2}) cos (y) = cos (y)

sin (\frac{π}{2}) cos (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 cos (y)

Multiply:

1 \cdot cos (y) = cos (y)

= cos (y)

cos (\frac{π}{2}) sin (y) = 0

cos (\frac{π}{2}) sin (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 sin (y)

Apply rule

0 \cdot a = 0

= 0

= cos (y) - 0

cos (y) - 0 = cos (y)

= cos (y)

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

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

= sec^{2} (y) - (\frac{sin ( y )}{cos ( y )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= sec^{2} (y) - \frac{sin ^{2} ( y )}{cos ^{2} ( y )}

Express with sin, cos

- \frac{sin ^{2} ( y )}{cos ^{2} ( y )} + sec^{2} (y)

Use the basic trigonometric identity:

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

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

Simplify

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

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

(\frac{1}{cos ( y )})^{2} = \frac{1}{cos ^{2} ( y )}

(\frac{1}{cos ( y )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cos ^{2} ( y )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( y )}

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

Apply rule

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

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

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

= \frac{1 - sin ^{2} ( y )}{cos ^{2} ( y )}

Rewrite using trig identities

\frac{1 - sin ^{2} ( y )}{cos ^{2} ( y )}

Use the Pythagorean identity:

1 = cos^{2} (x) + sin^{2} (x)

1 - sin^{2} (x) = cos^{2} (x)

= \frac{cos ^{2} ( y )}{cos ^{2} ( y )}

Apply rule

\frac{a}{a} = 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

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Is sec^2(y)-cot^2(pi/2-y)=1 ?
The answer to whether sec^2(y)-cot^2(pi/2-y)=1 is True