Is 1/(tan^2(x))=(cos^2(x))/(sin^2(x)) ?

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

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prove 1/(tan^2(x))=(cos^2(x))/(sin^2(x))

prove

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

True

Solution steps

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

Manipulating left 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

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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)

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