Is sec(2a)=(csc^2(a))/(cot^2(a)-1) ?

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

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Popular Trigonometry >

prove sec(2a)=(csc^2(a))/(cot^2(a)-1)

Solution

prove

sec (2 a) = \frac{csc ^{2} ( a )}{cot ^{2} ( a ) - 1}

Solution

True

Solution steps

sec (2 a) = \frac{csc ^{2} ( a )}{cot ^{2} ( a ) - 1}

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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} ( a ) + cos ^{2} ( a )}{sin ^{2} ( a )}

Multiply:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (a)

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sec (2 a)

sec (2 a)

sec (2 a)

We showed that the two sides could take the same form

\Rightarrow True

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