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

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

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prove sin(t)tan(t)=(1-cos^2(t))/(cos(t))

prove

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

True

Solution steps

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

Manipulating left side

sin (t) tan (t)

Express with sin, cos

sin (t) tan (t)

Use the basic trigonometric identity:

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

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

Simplify

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

sin (t) \frac{sin ( t )}{cos ( t )}

Multiply fractions:

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

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

sin (t) sin (t) = sin^{2} (t)

sin (t) sin (t)

Apply exponent rule:

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

sin (t) sin (t) = sin^{1 + 1} (t)

= sin^{1 + 1} (t)

Add the numbers:

1 + 1 = 2

= sin^{2} (t)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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