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

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

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

prove

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

True

Solution steps

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

Manipulating left side

1 + tan^{2} (t)

Express with sin, cos

1 + tan^{2} (t)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Convert element to fraction:

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

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

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

Multiply:

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

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

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

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

Rewrite using trig identities

\frac{cos ^{2} ( t ) + sin ^{2} ( t )}{cos ^{2} ( t )}

Use the Pythagorean identity:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ^{2} ( θ )}{cos ( θ )} = tan (θ) sin (θ)

cos^{2} (x) csc (x) sec (x) = cot (x)

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

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

sin^{2} (θ) - cos^{2} (θ) sin^{2} (θ) = sin^{4} (θ)

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