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

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

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

prove

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

True

Solution steps

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

Manipulating right side

tan (x) - cos (x) sin (x)

Express with sin, cos

tan (x) - cos (x) sin (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

sin (x) - cos (x) sin (x) cos (x)

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

cos (x) sin (x) cos (x)

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

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

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

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

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

Factor

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

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

Factor

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

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

Factor out common term

sin (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

sin^{2} (x) sin (x) = sin^{3} (x)

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

Apply exponent rule:

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

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

= sin^{2 + 1} (x)

Add the numbers:

2 + 1 = 3

= sin^{3} (x)

= \frac{sin ^{3} ( x )}{cos ( x )}

= \frac{sin ^{3} ( x )}{cos ( x )}

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

1 - \frac{sin ( x )}{1 + cos ( x )} = cos (x)

2 cos (3 x) = cot (3 x)

cos^{2} (θ) tan^{2} (θ) + cos^{2} (θ) = 1

\frac{sin ( θ ) csc ( θ )}{tan ( θ )} = cot (θ)

sec^{2} (θ) + csc (θ) = - sec^{2} (θ) csc (θ)

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