Is sin(x)((1+cot^2(x)))=csc(x) ?

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

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prove sin(x)((1+cot^2(x)))=csc(x)

prove

sin (x) ((1 + cot^{2} (x))) = csc (x)

True

Solution steps

sin (x) (1 + cot^{2} (x)) = csc (x)

Manipulating left side

sin (x) (1 + cot^{2} (x))

Express with sin, cos

(1 + cot^{2} (x)) sin (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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

Multiply:

1 \cdot sin^{2} (x) = sin^{2} (x)

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( x )}}

Simplify

\frac{1}{\frac{1}{c s c ( x )}}

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= csc (x)

csc (x)

csc (x)

We showed that the two sides could take the same form

\Rightarrow True

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

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

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

sin (3 θ) + sin (θ) = 2 sin (2 θ) cos (θ)

2 sin (x) \cdot cos (x) = sin (2 x)

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