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

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

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prove sin(θ)+(cos^2(θ))/(sin(θ))=csc(θ)

prove

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

True

Solution steps

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

Manipulating right side

csc (θ)

Express with sin, cos

csc (θ)

Use the basic trigonometric identity:

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

= \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

Rewrite using trig identities

\frac{1}{sin ( θ )}

Use the Pythagorean identity:

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

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

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

Expand

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

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

Apply the fraction rule:

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

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

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

Cancel

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

\frac{sin ^{2} ( θ )}{sin ( θ )}

Cancel the common factor:

sin (θ)

= sin (θ)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{cos ( - x ) tan ( - x )}{sin ( x )} = - 1

sin^{2} (20 0^{\circ}) = sin (20 0^{\circ}) sin (20 0^{\circ})

tan (θ) cot (θ) + sin^{2} (θ) + cos^{2} (θ) = 2

\frac{( 1 + tan ^{2} ( x ) )}{tan ^{2} ( x )} = csc^{2} (x)

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