Is 1csc(θ)-sin(θ)=cot(θ)cos(θ) ?

The answer to whether 1csc(θ)-sin(θ)=cot(θ)cos(θ) is True

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prove 1csc(θ)-sin(θ)=cot(θ)cos(θ)

prove

1 csc (θ) - sin (θ) = cot (θ) cos (θ)

True

Solution steps

1 \cdot csc (θ) - sin (θ) = cot (θ) cos (θ)

Manipulating left side

1 csc (θ) - sin (θ)

Express with sin, cos

- sin (θ) + 1 \cdot csc (θ)

Use the basic trigonometric identity:

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

= - sin (θ) + 1 \cdot \frac{1}{sin ( θ )}

Simplify

- sin (θ) + 1 \cdot \frac{1}{sin ( θ )} : \frac{- sin ^{2} ( θ ) + 1}{sin ( θ )}

- sin (θ) + 1 \cdot \frac{1}{sin ( θ )}

Multiply:

1 \cdot \frac{1}{sin ( θ )} = \frac{1}{sin ( θ )}

= - sin (θ) + \frac{1}{sin ( θ )}

Convert element to fraction:

sin (θ) = \frac{sin ( θ ) sin ( θ )}{sin ( θ )}

= - \frac{sin ( θ ) sin ( θ )}{sin ( θ )} + \frac{1}{sin ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{- sin ( θ ) sin ( θ ) + 1}{sin ( θ )}

- sin (θ) sin (θ) + 1 = - sin^{2} (θ) + 1

- sin (θ) sin (θ) + 1

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= - sin^{2} (θ) + 1

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

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

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

Rewrite using trig identities

\frac{1 - sin ^{2} ( θ )}{sin ( θ )}

Use the Pythagorean identity:

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

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

= \frac{cos ^{2} ( θ )}{sin ( θ )}

= \frac{cos ^{2} ( θ )}{sin ( θ )}

Rewrite using trig identities

= cos (θ) \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

cos (θ) cot (θ)

cos (θ) cot (θ)

= cot (θ) cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sec ( x )}{csc ^{2} ( x )} = sec (x) - cos (x)

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

sin (x) cos (x) tan (x) cot (x) = 1

(1 - cos (x)) = cos (x) - 1

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

Is 1csc(θ)-sin(θ)=cot(θ)cos(θ) ?
The answer to whether 1csc(θ)-sin(θ)=cot(θ)cos(θ) is True