Is (2-sec^2(x))/(sec^2(x))=cos(2x) ?

The answer to whether (2-sec^2(x))/(sec^2(x))=cos(2x) is True

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prove (2-sec^2(x))/(sec^2(x))=cos(2x)

prove

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

True

Solution steps

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

Manipulating left side

\frac{2 - sec ^{2} ( x )}{sec ^{2} ( x )}

Express with sin, cos

\frac{2 - sec ^{2} ( x )}{sec ^{2} ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

= \frac{( 2 - \frac{1}{c o s ^{2} ( x )} ) cos ^{2} ( x )}{1}

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

= \frac{\frac{2 c o s ^{2} ( x ) - 1}{c o s ^{2} ( x )} cos ^{2} ( x )}{1}

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

= 2 cos^{2} (x) - 1

= 2 cos^{2} (x) - 1

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

Rewrite using trig identities

- 1 + 2 cos^{2} (x)

Use the Double Angle identity:

2 cos^{2} (x) - 1 = cos (2 x)

= cos (2 x)

= cos (2 x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{cos ( β )}{1 - sin ( β )} = sec (β) + tan (β)

csc^{2} (x) = \frac{tan ^{2} ( x ) + 1}{tan ^{2} ( x )}

2 cot (2 θ) = cot (θ) - tan (θ)

7 cos^{2} (b) - 7 sin^{2} (b) = 14 cos^{2} (b) - 7

1 + tan (x) \cdot tan (2 x) = sec (2 x)

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