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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Expand

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

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

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

Apply the distributive law:

a (b - c) = ab - a c

a = tan^{2} (x), b = 1, c = cos^{2} (x)

= tan^{2} (x) \cdot 1 - tan^{2} (x) cos^{2} (x)

= 1 \cdot tan^{2} (x) - tan^{2} (x) cos^{2} (x)

Multiply:

1 \cdot tan^{2} (x) = tan^{2} (x)

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

= tan^{2} (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)

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

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

We showed that the two sides could take the same form

\Rightarrow True

sin^{4} (x) - cos^{4} (x) = 2 sin^{2} (x) - 1

(sec (θ) - 1) (sec (θ) + 1) = tan^{2} (θ)

1 + tan^{2} (x) = sec^{2} (x)

sin^{3} (2 x) = (\frac{1}{2} sin (2 x)) (1 - cos (4 x))

cos (θ) tan (θ) = sin (θ)

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