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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

sec^{2} (\frac{π}{2} - x) - 1

Rewrite using trig identities

sec (\frac{π}{2} - x)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( \frac{π}{2} - x )}

Use the Angle Difference identity:

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

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

Simplify

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

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

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

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

cos (\frac{π}{2}) cos (x) = 0

cos (\frac{π}{2}) cos (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 cos (x)

Apply rule

0 \cdot a = 0

= 0

sin (\frac{π}{2}) sin (x) = sin (x)

sin (\frac{π}{2}) sin (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 sin (x)

Multiply:

1 \cdot sin (x) = sin (x)

= sin (x)

= 0 + sin (x)

0 + sin (x) = sin (x)

= sin (x)

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

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

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

(\frac{1}{sin ( x )})^{2} = \frac{1}{sin ^{2} ( x )}

(\frac{1}{sin ( x )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sin ^{2} ( x )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( x )}

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

Manipulating right side

cot^{2} (x)

Express with sin, cos

cot^{2} (x)

Use the basic trigonometric identity:

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

= (\frac{cos ( x )}{sin ( x )})^{2}

Simplify

(\frac{cos ( x )}{sin ( x )})^{2} : \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

(\frac{cos ( x )}{sin ( x )})^{2}

Apply exponent rule:

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

= \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

= \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

= \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

Rewrite using trig identities

\frac{cos ^{2} ( x )}{sin ^{2} ( x )}

Use the Pythagorean identity:

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

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

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

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

Expand

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

\frac{1 - sin ^{2} ( x )}{sin ^{2} ( x )}

Apply the fraction rule:

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

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

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

Apply rule

\frac{a}{a} = 1

\frac{sin ^{2} ( x )}{sin ^{2} ( x )} = 1

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

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

We showed that the two sides could take the same form

\Rightarrow True

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