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

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

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

prove

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

True

Solution steps

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

Manipulating left side

tan^{2} (x) cos^{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

\frac{tan ^{2} ( x ) + 1}{1 - sin ^{2} ( x )} = sec^{4} (x)

tan (A) = \frac{sec ( A )}{csc ( A )}

csc (- x) = - csc (x)

sin (β) + cos (β) cot (β) = csc (β)

cos (\frac{3 π}{2} + x) = - cos (x)

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