Is cos(θ)sec(pi/2-θ)=cot(θ) ?

The answer to whether cos(θ)sec(pi/2-θ)=cot(θ) is True

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prove cos(θ)sec(pi/2-θ)=cot(θ)

prove

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

True

Solution steps

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

Manipulating left side

cos (θ) sec (\frac{π}{2} - θ)

Rewrite using trig identities

sec (\frac{π}{2} - θ)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( \frac{π}{2} - θ )}

Use the Angle Difference identity:

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

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

Simplify

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

\frac{1}{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{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

= cos (θ) \frac{1}{sin ( θ )}

Simplify

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

cos (θ) \frac{1}{sin ( θ )}

Multiply fractions:

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

= \frac{1 \cdot cos ( θ )}{sin ( θ )}

Multiply:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

= cot (θ)

We showed that the two sides could take the same form

\Rightarrow True

4 cos^{4} (x) - 4 sin^{4} (x) = 4 cos (2 x)

csc (x) (1 + sin (x)) = 1 + csc (x)

sin (x + y) sin (x - y) = sin (2 x) - sin (2 y)

csc^{2} (x) = 1 - cot^{2} (x)

cos^{2} (θ) = 1 + sin^{2} (θ)

Is cos(θ)sec(pi/2-θ)=cot(θ) ?
The answer to whether cos(θ)sec(pi/2-θ)=cot(θ) is True