Is csc(Θ)+sin(-Θ)=(cos^2(Θ))/(sin(Θ)) ?

The answer to whether csc(Θ)+sin(-Θ)=(cos^2(Θ))/(sin(Θ)) is True

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prove csc(Θ)+sin(-Θ)=(cos^2(Θ))/(sin(Θ))

prove

csc (Θ) + sin (- Θ) = \frac{cos ^{2} ( Θ )}{sin ( Θ )}

True

Solution steps

csc (Θ) + sin (- Θ) = \frac{cos ^{2} ( Θ )}{sin ( Θ )}

Manipulating left side

csc (Θ) + sin (- Θ)

Use the negative angle identity:

sin (- x) = - sin (x)

= 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 )}

We showed that the two sides could take the same form

\Rightarrow True

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

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

csc (A) = \frac{cos ( 2 A )}{sin ( A )} + \frac{sin ( 2 A )}{cos ( A )}

sin (θ) csc (θ) + cot (θ) sin (θ) = 1 + cos (θ)

tan (x) cot (x) = csc (x) sec (x)

Is csc(Θ)+sin(-Θ)=(cos^2(Θ))/(sin(Θ)) ?
The answer to whether csc(Θ)+sin(-Θ)=(cos^2(Θ))/(sin(Θ)) is True