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

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

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

prove

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

True

Solution steps

\frac{cos ^{2} ( θ )}{sin ( 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)

We showed that the two sides could take the same form

\Rightarrow True

cos (4 t) = 1 - 8 sin^{2} (t) cos^{2} (t)

sin (10 5^{\circ}) = sin (6 0^{\circ} + 4 5^{\circ})

cos (θ) = sin (θ + \frac{π}{2})

cos^{3} (x) = sin (x)

sin (θ) = - \frac{5}{12}

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