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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

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

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

Apply exponent rule:

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

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

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

Apply the fraction rule:

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (x)

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Apply rule

\frac{a}{1} = a

= cos^{2} (x)

= cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{1 - cot ( - θ )}{cot ( - θ )} = - tan (θ) - 1

cos (6 0^{\circ}) = 2 cos^{2} (3 0^{\circ}) - 1

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

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

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

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