Is (cot(x))/(tan(x))= 1/(tan^2(x)) ?

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

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prove (cot(x))/(tan(x))= 1/(tan^2(x))

prove

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

True

Solution steps

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

Manipulating left side

\frac{cot ( x )}{tan ( x )}

Express with sin, cos

\frac{cot ( x )}{tan ( x )}

Use the basic trigonometric identity:

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

= \frac{\frac{c o s ( x )}{s i n ( x )}}{tan ( x )}

Use the basic trigonometric identity:

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

= \frac{\frac{c o s ( x )}{s i n ( x )}}{\frac{s i n ( x )}{c o s ( x )}}

Simplify

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

\frac{\frac{c o s ( x )}{s i n ( x )}}{\frac{s i n ( x )}{c o s ( x )}}

Divide fractions:

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

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

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

cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

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

Manipulating right side

\frac{1}{tan ^{2} ( x )}

Express with sin, cos

\frac{1}{tan ^{2} ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

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

Apply the fraction rule:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

2 - sec^{2} (θ) = 1 - tan^{2} (θ)

\frac{tan ( θ ) + sec ^{2} ( θ ) - sec ( θ )}{sec ( θ )} = tan^{2} (θ) + sin (θ)

sin (\frac{9 π}{4}) = sin (\frac{3}{4} π + 2 π)

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

\frac{1 - tan ^{2} ( A )}{1 + tan ^{2} ( A )} = cos (2 A)

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