Is cos(x)tan^3(x)=sin(x)tan^2(x) ?

The answer to whether cos(x)tan^3(x)=sin(x)tan^2(x) is True

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prove cos(x)tan^3(x)=sin(x)tan^2(x)

prove

cos (x) tan^{3} (x) = sin (x) tan^{2} (x)

True

Solution steps

cos (x) tan^{3} (x) = sin (x) tan^{2} (x)

Manipulating left side

cos (x) tan^{3} (x)

Express with sin, cos

cos (x) tan^{3} (x)

Use the basic trigonometric identity:

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

= cos (x) (\frac{sin ( x )}{cos ( x )})^{3}

Simplify

cos (x) (\frac{sin ( x )}{cos ( x )})^{3} : \frac{sin ^{3} ( x )}{cos ^{2} ( x )}

cos (x) (\frac{sin ( x )}{cos ( x )})^{3}

(\frac{sin ( x )}{cos ( x )})^{3} = \frac{sin ^{3} ( x )}{cos ^{3} ( x )}

(\frac{sin ( x )}{cos ( x )})^{3}

Apply exponent rule:

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

= \frac{sin ^{3} ( x )}{cos ^{3} ( x )}

= \frac{sin ^{3} ( x )}{cos ^{3} ( x )} cos (x)

Multiply fractions:

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

= \frac{sin ^{3} ( x ) cos ( x )}{cos ^{3} ( x )}

Cancel the common factor:

cos (x)

= \frac{sin ^{3} ( x )}{cos ^{2} ( x )}

= \frac{sin ^{3} ( x )}{cos ^{2} ( x )}

= \frac{sin ^{3} ( x )}{cos ^{2} ( x )}

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

Use the basic trigonometric identity:

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

sin (x) tan (x) tan (x)

Simplify

sin (x) tan (x) tan (x) : tan^{2} (x) sin (x)

sin (x) tan (x) tan (x)

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

Is cos(x)tan^3(x)=sin(x)tan^2(x) ?
The answer to whether cos(x)tan^3(x)=sin(x)tan^2(x) is True