Is (cot^2(x)+1)/(cot^2(x)-1)=sec(2x) ?

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

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Popular Trigonometry >

prove (cot^2(x)+1)/(cot^2(x)-1)=sec(2x)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Simplify

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

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

Apply exponent rule:

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

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

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

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

Divide fractions:

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

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

Cancel the common factor:

sin^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

cos (x) = \frac{1}{sec ( x )}

\frac{1}{\frac{1}{s e c ( 2 x )}}

Simplify

\frac{1}{\frac{1}{s e c ( 2 x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{sec ( 2 x )}{1}

Apply rule

\frac{a}{1} = a

= sec (2 x)

sec (2 x)

sec (2 x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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