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

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

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prove 1/(cos^2(x))=1+tan^2(x)

prove

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

True

Solution steps

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

Manipulating right side

1 + tan^{2} (x)

Express with sin, cos

1 + tan^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Convert element to fraction:

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

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

Multiply:

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

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

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

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

Rewrite using trig identities

\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 )}

We showed that the two sides could take the same form

\Rightarrow True

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

sin (θ - \frac{3 π}{2}) = cos (θ)

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

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

sin (27 0^{\circ} + x) = - cos (x)

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