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

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

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

prove

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

True

Solution steps

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

Manipulating right side

csc (x)

Express with sin, cos

csc (x)

Use the basic trigonometric identity:

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

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

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

Rewrite using trig identities

\frac{1}{sin ( x )}

Use the Pythagorean identity:

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

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

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

Expand

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

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

Apply the fraction rule:

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

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

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

Cancel

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

\frac{sin ^{2} ( x )}{sin ( x )}

Cancel the common factor:

sin (x)

= sin (x)

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

sec (θ) = sin (θ) (tan (θ) + cot (θ))

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

sin (3 a) = 3 sin (a) - 4 sin^{3} (a)

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