Is (cot^2(B)-cos^2(B))/(csc^2(B)-1)=cos^2(B) ?

The answer to whether (cot^2(B)-cos^2(B))/(csc^2(B)-1)=cos^2(B) is True

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prove (cot^2(B)-cos^2(B))/(csc^2(B)-1)=cos^2(B)

prove

\frac{cot ^{2} ( B ) - cos ^{2} ( B )}{csc ^{2} ( B ) - 1} = cos^{2} (B)

True

Solution steps

\frac{cot ^{2} ( B ) - cos ^{2} ( B )}{csc ^{2} ( B ) - 1} = cos^{2} (B)

Manipulating left side

\frac{cot ^{2} ( B ) - cos ^{2} ( B )}{csc ^{2} ( B ) - 1}

Express with sin, cos

\frac{- cos ^{2} ( B ) + cot ^{2} ( B )}{- 1 + csc ^{2} ( B )}

Use the basic trigonometric identity:

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

= \frac{- cos ^{2} ( B ) + ( \frac{c o s ( B )}{s i n ( B )} ) ^{2}}{- 1 + csc ^{2} ( B )}

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= \frac{- cos ^{2} ( B ) + ( \frac{c o s ( B )}{s i n ( B )} ) ^{2}}{- 1 + ( \frac{1}{s i n ( B )} ) ^{2}}

Simplify

\frac{- cos ^{2} ( B ) + ( \frac{c o s ( B )}{s i n ( B )} ) ^{2}}{- 1 + ( \frac{1}{s i n ( B )} ) ^{2}} : cos^{2} (B)

\frac{- cos ^{2} ( B ) + ( \frac{c o s ( B )}{s i n ( B )} ) ^{2}}{- 1 + ( \frac{1}{s i n ( B )} ) ^{2}}

(\frac{1}{sin ( B )})^{2} = \frac{1}{sin ^{2} ( B )}

(\frac{1}{sin ( B )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sin ^{2} ( B )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( B )}

= \frac{- cos ^{2} ( B ) + ( \frac{c o s ( B )}{s i n ( B )} ) ^{2}}{- 1 + \frac{1}{s i n ^{2} ( B )}}

Apply exponent rule:

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

= \frac{- cos ^{2} ( B ) + \frac{c o s ^{2} ( B )}{s i n ^{2} ( B )}}{- 1 + \frac{1}{s i n ^{2} ( B )}}

Join

- 1 + \frac{1}{sin ^{2} ( B )} : \frac{- sin ^{2} ( B ) + 1}{sin ^{2} ( B )}

- 1 + \frac{1}{sin ^{2} ( B )}

Convert element to fraction:

1 = \frac{1 sin ^{2} ( B )}{sin ^{2} ( B )}

= - \frac{1 \cdot sin ^{2} ( B )}{sin ^{2} ( B )} + \frac{1}{sin ^{2} ( B )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot sin ^{2} ( B ) + 1}{sin ^{2} ( B )}

Multiply:

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

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

= \frac{- cos ^{2} ( B ) + \frac{c o s ^{2} ( B )}{s i n ^{2} ( B )}}{\frac{- s i n ^{2} ( B ) + 1}{s i n ^{2} ( B )}}

Join

- cos^{2} (B) + \frac{cos ^{2} ( B )}{sin ^{2} ( B )} : \frac{- cos ^{2} ( B ) sin ^{2} ( B ) + cos ^{2} ( B )}{sin ^{2} ( B )}

- cos^{2} (B) + \frac{cos ^{2} ( B )}{sin ^{2} ( B )}

Convert element to fraction:

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

= - \frac{cos ^{2} ( B ) sin ^{2} ( B )}{sin ^{2} ( B )} + \frac{cos ^{2} ( B )}{sin ^{2} ( B )}

Since the denominators are equal, combine the fractions:

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

= \frac{- cos ^{2} ( B ) sin ^{2} ( B ) + cos ^{2} ( B )}{sin ^{2} ( B )}

= \frac{\frac{- c o s ^{2} ( B ) s i n ^{2} ( B ) + c o s ^{2} ( B )}{s i n ^{2} ( B )}}{\frac{- s i n ^{2} ( B ) + 1}{s i n ^{2} ( B )}}

Divide fractions:

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

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

Cancel the common factor:

sin^{2} (B)

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

Factor out common term

cos^{2} (B)

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

Cancel the common factor:

- sin^{2} (B) + 1

= cos^{2} (B)

= cos^{2} (B)

= cos^{2} (B)

We showed that the two sides could take the same form

\Rightarrow True

Is (cot^2(B)-cos^2(B))/(csc^2(B)-1)=cos^2(B) ?
The answer to whether (cot^2(B)-cos^2(B))/(csc^2(B)-1)=cos^2(B) is True