Is sin(θ)(1+cot^2(θ))=csc(θ) ?

The answer to whether sin(θ)(1+cot^2(θ))=csc(θ) is True

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prove sin(θ)(1+cot^2(θ))=csc(θ)

prove

sin (θ) (1 + cot^{2} (θ)) = csc (θ)

True

Solution steps

sin (θ) (1 + cot^{2} (θ)) = csc (θ)

Manipulating left side

sin (θ) (1 + cot^{2} (θ))

Express with sin, cos

(1 + cot^{2} (θ)) sin (θ)

Use the basic trigonometric identity:

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

= (1 + (\frac{cos ( θ )}{sin ( θ )})^{2}) sin (θ)

Simplify

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

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

Apply exponent rule:

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

= sin (θ) (\frac{cos ^{2} ( θ )}{sin ^{2} ( θ )} + 1)

Join

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

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

Convert element to fraction:

1 = \frac{1 sin ^{2} ( θ )}{sin ^{2} ( θ )}

= \frac{1 \cdot sin ^{2} ( θ )}{sin ^{2} ( θ )} + \frac{cos ^{2} ( θ )}{sin ^{2} ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot sin ^{2} ( θ ) + cos ^{2} ( θ )}{sin ^{2} ( θ )}

Multiply:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (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

csc (x) - sin (x) = csc (x) cot (x)

5 + cot^{2} (x) = 4 + csc^{2} (x)

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

sin^{2} (2 x) = \frac{1 - cos ( 4 x )}{2}

cot (2 θ) = \frac{csc ( θ ) - 2 sin ( θ )}{2 cos ( θ )}

Is sin(θ)(1+cot^2(θ))=csc(θ) ?
The answer to whether sin(θ)(1+cot^2(θ))=csc(θ) is True