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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Factor

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

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

Factor out common term

tan^{2} (b)

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Express with sin, cos

cos^{2} (b) tan^{2} (b)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (b)

= sin^{2} (b)

= sin^{2} (b)

= sin^{2} (b)

We showed that the two sides could take the same form

\Rightarrow True

2 cos (x) sin (x) = sin (2 x)

\frac{cot ( x ) \cdot sec ( x )}{csc ( x )} = 1

sin (x) + tan (x) = tan (x) (1 + cos (x))

3 sin (x) cos (x) = 3 sin^{2} (x) cot (x)

sin (\frac{3 π}{2} - 0) = - cos (0)

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