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

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

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

prove

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

True

Solution steps

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

Manipulating right side

sec (x) tan (x)

Express with sin, cos

sec (x) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

csc (x) tan (x) + sec (x) = 2 sec (x)

csc (x) (sin (x) + tan (x)) = 1 + sec (x)

\frac{1}{tan ( 7 0 ^{\circ} )} = cot (7 0^{\circ})

\frac{sec ( A )}{sin ( A )} - \frac{sin ( A )}{cos ( A )} = cot (A)

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