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

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

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

prove

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{1}{1 + ( \frac{c o s ( x )}{s i n ( x )} ) ^{2}}

Simplify

\frac{1}{1 + ( \frac{c o s ( x )}{s i n ( x )} ) ^{2}} : \frac{sin ^{2} ( x )}{sin ^{2} ( x ) + cos ^{2} ( x )}

\frac{1}{1 + ( \frac{c o s ( x )}{s i n ( x )} ) ^{2}}

Apply exponent rule:

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

= \frac{1}{1 + \frac{c o s ^{2} ( x )}{s i n ^{2} ( x )}}

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{1}{\frac{s i n ^{2} ( x ) + c o s ^{2} ( x )}{s i n ^{2} ( x )}}

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Apply rule

\frac{a}{1} = a

= sin^{2} (x)

= sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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