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

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

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prove cos^2(x)=((csc^2(x)-1))/(csc^2(x))

prove

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

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{- 1 + ( \frac{1}{s i n ( x )} ) ^{2}}{( \frac{1}{s i n ( x )} ) ^{2}}

Simplify

\frac{- 1 + ( \frac{1}{s i n ( x )} ) ^{2}}{( \frac{1}{s i n ( x )} ) ^{2}} : - sin^{2} (x) + 1

\frac{- 1 + ( \frac{1}{s i n ( x )} ) ^{2}}{( \frac{1}{s i n ( x )} ) ^{2}}

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{- 1 + ( \frac{1}{s i n ( x )} ) ^{2}}{\frac{1}{s i n ^{2} ( x )}}

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{- 1 + \frac{1}{s i n ^{2} ( x )}}{\frac{1}{s i n ^{2} ( x )}}

Apply the fraction rule:

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

= \frac{( - 1 + \frac{1}{s i n ^{2} ( x )} ) sin ^{2} ( x )}{1}

Join

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

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

Convert element to fraction:

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

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

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

Multiply:

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

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

= \frac{\frac{- s i n ^{2} ( x ) + 1}{s i n ^{2} ( x )} sin ^{2} ( x )}{1}

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (x)

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

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

= 1 - sin^{2} (x)

Rewrite using trig identities

1 - sin^{2} (x)

Use the Pythagorean identity:

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

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

= cos^{2} (x)

= cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

cos (30 0^{\circ}) = 1 - 2 sin^{2} (15 0^{\circ})

csc^{2} (x) + 3 cot^{2} (x) - 5 = 4 (cot (x) - 1)

(3) \frac{( cos ( 2 z ) ) ^{2}}{2} = \frac{3 cos ( 4 z )}{4}

- 2 sin^{2} (x) + cos (x) + 1 = 0

\frac{2 cot ( u )}{csc ^{2} ( u ) - 2} = tan (2 u)

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