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

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

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

prove

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

True

Solution steps

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

Manipulating left side

sin (x) tan (x)

Express with sin, cos

sin (x) tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

sec (u) + tan (u) = \frac{cos ( u )}{1 - sin ( u )}

cos (x - \frac{π}{6}) - cos (x + \frac{π}{6}) = sin (x)

cos (\frac{3 π}{2} + θ) = sin (θ)

csc^{2} (θ) - cos^{2} (θ) csc^{2} (θ) = 1

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

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