Is csc(θ)-sin(θ)=cos(θ)cot(θ) ?

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

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prove csc(θ)-sin(θ)=cos(θ)cot(θ)

prove

csc (θ) - sin (θ) = cos (θ) cot (θ)

True

Solution steps

csc (θ) - sin (θ) = cos (θ) cot (θ)

Manipulating left side

csc (θ) - sin (θ)

Express with sin, cos

csc (θ) - sin (θ)

Use the basic trigonometric identity:

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

= \frac{1}{sin ( θ )} - sin (θ)

Simplify

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

\frac{1}{sin ( θ )} - sin (θ)

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 - sin (θ) sin (θ) = 1 - sin^{2} (θ)

1 - sin (θ) sin (θ)

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)

= 1 - sin^{2} (θ)

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

= \frac{1 - sin ^{2} ( θ )}{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)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sec ( x )}{tan ( x ) + cot ( x )} = sin (x)

cos (θ) = \frac{1}{sec ( θ )}

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

4 (sin^{6} (x) + cos^{6} (x)) = 4 - 3 sin^{2} (2 x)

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

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