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

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

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

prove

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

True

Solution steps

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

Manipulating left side

tan^{2} (θ) - tan^{2} (θ) sin^{2} (θ)

Factor

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

tan^{2} (θ) - sin^{2} (θ) tan^{2} (θ)

Factor out common term

tan^{2} (θ)

= tan^{2} (θ) (1 - sin^{2} (θ))

= (1 - sin^{2} (θ)) tan^{2} (θ)

Rewrite using trig identities

(1 - sin^{2} (θ)) tan^{2} (θ)

Use the Pythagorean identity:

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

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

= cos^{2} (θ) tan^{2} (θ)

= cos^{2} (θ) tan^{2} (θ)

Express with sin, cos

cos^{2} (θ) tan^{2} (θ)

Use the basic trigonometric identity:

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

= 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)

= sin^{2} (θ)

= sin^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

(1 + sin (z)) (1 - sin (z)) = \frac{1}{sec ^{2} ( z )}

\frac{tan ( x )}{sec ( x ) + 1} = csc (x) - cot (x)

cos^{2} (x) = \frac{1}{sec ^{2} ( x )}

\frac{( 1 + tan ( θ ) )}{1 + cot ( θ )} = tan (θ)

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