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

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

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

prove

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

True

Solution steps

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

Manipulating left side

1 + cot^{2} (x)

Express with sin, cos

1 + cot^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

= 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 )}

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{1 + tan ( x )}{1 + cot ( x )} = tan (x)

tan (θ) cot (θ) - sin^{2} (θ) = cos^{2} (θ)

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

sec (θ) = \frac{1}{cos ( θ )}

csc (2 θ) = \frac{csc ( θ )}{2 cos ( θ )}

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