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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Expand

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

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Multiply:

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

= tan^{2} (B) - 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)

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

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{cos ( 2 x ) + 1}{sin ( 2 x )} = cot (x)

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

- tan (t) - \frac{cos ( t )}{sin ( t ) - 1} = sec (x)

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

cos (\frac{x}{2}) = \frac{( 1 + cos ( x ) )}{2}

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