Is 1/(1-sin^2(θ))=1+tan^2(θ) ?

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

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prove 1/(1-sin^2(θ))=1+tan^2(θ)

prove

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

True

Solution steps

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

Manipulating right 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

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

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

Add similar elements:

- sin^{2} (θ) + sin^{2} (θ) = 0

= \frac{1}{1 - sin ^{2} ( θ )}

= \frac{1}{1 - sin ^{2} ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

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cos (θ) = sin (\frac{π}{2} + θ)

1 - 8 sin^{2} (θ) cos^{2} (θ) = cos (4 θ)

\frac{cos ( b )}{sin ( b )} \cdot sin (b) = cos (b)

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