Is (sin(θ))/(1-sin^2(θ))=sec(θ)tan(θ) ?

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

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prove (sin(θ))/(1-sin^2(θ))=sec(θ)tan(θ)

prove

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

True

Solution steps

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

Manipulating right side

sec (θ) tan (θ)

Express with sin, cos

sec (θ) tan (θ)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( θ )} tan (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

\frac{1}{cos ( θ )} \cdot \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

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

Multiply:

1 \cdot sin (θ) = sin (θ)

= \frac{sin ( θ )}{cos ( θ ) cos ( θ )}

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

= \frac{sin ( θ )}{cos ^{2} ( θ )}

= \frac{sin ( θ )}{cos ^{2} ( θ )}

= \frac{sin ( θ )}{cos ^{2} ( θ )}

Rewrite using trig identities

\frac{sin ( θ )}{cos ^{2} ( θ )}

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

sin^{2} (θ) cot^{2} (θ) = 1

tan^{2} (x) = csc^{2} (x) tan^{2} (x) - 1

(1 - tan (θ))^{2} = sec^{2} (θ) - 2 tan (θ)

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

4 sin (\frac{x}{4}) cos (\frac{x}{4}) cos (\frac{x}{2}) = sin (x)

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