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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

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

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

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (x)

= cos^{2} (x)

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

Cancel the common factor:

cos (x)

= cos (x)

= cos (x)

= cos (x)

We showed that the two sides could take the same form

\Rightarrow True

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

(sin (x) - 1) (tan (x) + sec (x)) = - cos (x)

cos (a - b) - cos (a + b) = 2 sin (a) sin (b)

tan (x) + cot (2 x) = csc (2 x)

tan (θ) sin (θ) = sec (θ) - cos (θ)

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