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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Factor

tan^{2} (x) - sin^{2} (x) tan^{2} (x) : tan^{2} (x) (1 - sin^{2} (x))

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

Factor out common term

tan^{2} (x)

= tan^{2} (x) (1 - sin^{2} (x))

= (1 - sin^{2} (x)) tan^{2} (x)

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

= 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

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

1 + tan^{2} (6 0^{\circ}) = sec^{2} (6 0^{\circ})

1 + cos (θ) = sin (θ)

sin (x) = sin (x + 2 π)

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

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