Is prove-cos^2(x)tan^2(x)=-sin^2(x) ?

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

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

prove

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

True

Solution steps

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

Manipulating left side

- cos^{2} (x) tan^{2} (x)

Express with sin, cos

- cos^{2} (x) tan^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

= - sin^{2} (x)

= - sin^{2} (x)

= - sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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

2 cos^{2} (x) - cos (x) - 1 = 0

cos (θ) = cot (θ) sin^{2} (θ) csc (θ)

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

(1 - cos^{2} (A)) (1 + cot^{2} (A)) = 1

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