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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Join

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

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

Multiply:

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

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

cos^{2} (x) - 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

cot (x) tan (x) = csc (x) sec (x)

sec (x) - cos (x) = (1 - cos (x)) (1 + sec (x))

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

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

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

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