Is sin(θ)+sin(θ)cot^2(θ)=csc(θ) ?

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

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prove sin(θ)+sin(θ)cot^2(θ)=csc(θ)

prove

sin (θ) + sin (θ) cot^{2} (θ) = csc (θ)

True

Solution steps

sin (θ) + sin (θ) cot^{2} (θ) = csc (θ)

Manipulating left side

sin (θ) + sin (θ) cot^{2} (θ)

Express with sin, cos

sin (θ) + cot^{2} (θ) sin (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

(\frac{cos ( θ )}{sin ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (θ)

= \frac{cos ^{2} ( θ )}{sin ( θ )}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

sin (θ) sin (θ) + cos^{2} (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)

= sin^{2} (θ) + cos^{2} (θ)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( θ )}}

Simplify

\frac{1}{\frac{1}{c s c ( θ )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{csc ( θ )}{1}

Apply rule

\frac{a}{1} = a

= csc (θ)

csc (θ)

csc (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{( cos ( x ) sec ( x ) )}{tan ( x )} = cot (x)

+ sin (z + w) = sin (z) cos (w) + cos (z) sin (w)

(sin^{2} (x) + cos^{2} (x))^{5} = 1

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

csc (θ) = \frac{sec ( θ )}{tan ( θ )}

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