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

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

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prove 1-tan^2(a)=cos(2a)sec^2(a)

prove

1 - tan^{2} (a) = cos (2 a) sec^{2} (a)

True

Solution steps

1 - tan^{2} (a) = cos (2 a) sec^{2} (a)

Manipulating left side

1 - tan^{2} (a)

Express with sin, cos

1 - tan^{2} (a)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Use the basic trigonometric identity:

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

= cos (2 a) sec^{2} (a)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ( x ) - cos ( x )}{sin ( x )} = 1 - cot (x)

\frac{( sec ( θ ) - 1 )}{( 1 - cos ( θ ) )} = sec (θ)

\frac{sin ( a ) - 2 sin ^{3} ( a )}{2 cos ^{3} ( a ) - cos ( a )} = tan (a)

cot (u - v) = \frac{cot ( u ) cot ( v ) + 1}{cot ( v ) - cot ( u )}

cos (0) tan (0) = sin (0)

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