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

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

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

prove

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

True

Solution steps

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

Manipulating right side

sin^{2} (θ) tan^{2} (θ)

Rewrite using trig identities

sin^{2} (θ) tan^{2} (θ)

Use the Pythagorean identity:

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

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

= (1 - cos^{2} (θ)) tan^{2} (θ)

= (1 - cos^{2} (θ)) tan^{2} (θ)

Expand

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

(1 - cos^{2} (θ)) tan^{2} (θ)

= tan^{2} (θ) (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Multiply:

1 \cdot tan^{2} (θ) = tan^{2} (θ)

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

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

Rewrite using trig identities

tan^{2} (θ) - cos^{2} (θ) tan^{2} (θ)

Use the basic trigonometric identity:

tan (θ) = \frac{sin ( θ )}{cos ( θ )}

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

Simplify

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

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

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

(\frac{sin ( θ )}{cos ( θ )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (θ)

= sin^{2} (θ)

= tan^{2} (θ) - sin^{2} (θ)

= tan^{2} (θ) - sin^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

cos^{2} (2 u) - sin^{2} (2 u) = cos (4 u)

tan (u) cot (u) - cos^{2} (u) = sin^{2} (u)

\frac{1 - sin ^{2} ( x )}{csc ^{2} ( x ) - 1} = sin^{2} (x)

cos (2 θ) = 1 - 2 sin^{2} (θ)

sin (x) = 1 - cos (x)

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