Is cos(2a)=-1+2/(1+tan^2(a)) ?

The answer to whether cos(2a)=-1+2/(1+tan^2(a)) is True

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prove cos(2a)=-1+2/(1+tan^2(a))

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

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

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

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

Apply the fraction rule:

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

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

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

Convert element to fraction:

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

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

Multiply:

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

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

Expand

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a

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

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

Add similar elements:

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Apply rule

\frac{a}{1} = a

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

Use the Double Angle identity:

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

= cos (2 a)

= cos (2 a)

We showed that the two sides could take the same form

\Rightarrow True

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Is cos(2a)=-1+2/(1+tan^2(a)) ?
The answer to whether cos(2a)=-1+2/(1+tan^2(a)) is True