Is 1+tan^2(θ)= 1/(cos^2(θ)) ?

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

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prove 1+tan^2(θ)= 1/(cos^2(θ))

prove

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

True

Solution steps

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

Manipulating left side

1 + tan^{2} (θ)

Express with sin, cos

1 + tan^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

1 + (\frac{sin ( θ )}{cos ( θ )})^{2}

Apply exponent rule:

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

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

Convert element to fraction:

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

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

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

Multiply:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= \frac{1}{cos ^{2} ( θ )}

= \frac{1}{cos ^{2} ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

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

sin (θ + π) = - sin (θ)

tan^{2} (x) sec^{2} (x) - tan^{2} (x) = tan^{4} (x)

sin^{4} (x) + cos^{4} (x) = \frac{3}{4} + \frac{1}{4} cos (4 x)

sin^{2} (x) + 3 cos^{2} (x) = 1 + 2 cos^{2} (x)

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