Is sin(x)cos(x)+sin^3(x)sec(x)=tan(x) ?

The answer to whether sin(x)cos(x)+sin^3(x)sec(x)=tan(x) is True

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prove sin(x)cos(x)+sin^3(x)sec(x)=tan(x)

prove

sin (x) cos (x) + sin^{3} (x) sec (x) = tan (x)

True

Solution steps

sin (x) cos (x) + sin^{3} (x) sec (x) = tan (x)

Manipulating left side

sin (x) cos (x) + sin^{3} (x) sec (x)

Express with sin, cos

cos (x) sin (x) + sec (x) sin^{3} (x)

Use the basic trigonometric identity:

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

= cos (x) sin (x) + \frac{1}{cos ( x )} sin^{3} (x)

Simplify

cos (x) sin (x) + \frac{1}{cos ( x )} sin^{3} (x) : \frac{cos ^{2} ( x ) sin ( x ) + sin ^{3} ( x )}{cos ( x )}

cos (x) sin (x) + \frac{1}{cos ( x )} sin^{3} (x)

\frac{1}{cos ( x )} sin^{3} (x) = \frac{sin ^{3} ( x )}{cos ( x )}

\frac{1}{cos ( x )} sin^{3} (x)

Multiply fractions:

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

= \frac{1 \cdot sin ^{3} ( x )}{cos ( x )}

Multiply:

1 \cdot sin^{3} (x) = sin^{3} (x)

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

cos (x) sin (x) cos (x) + sin^{3} (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)

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

= \frac{cos ^{2} ( x ) sin ( x ) + sin ^{3} ( x )}{cos ( x )}

= \frac{cos ^{2} ( x ) sin ( x ) + sin ^{3} ( x )}{cos ( x )}

= \frac{sin ^{3} ( x ) + cos ^{2} ( x ) sin ( x )}{cos ( x )}

Factor

\frac{sin ^{3} ( x ) + cos ^{2} ( x ) sin ( x )}{cos ( x )} : \frac{sin ( x ) ( sin ^{2} ( x ) + cos ^{2} ( x ) )}{cos ( x )}

\frac{sin ^{3} ( x ) + cos ^{2} ( x ) sin ( x )}{cos ( x )}

Factor

sin^{3} (x) + cos^{2} (x) sin (x) : sin (x) (sin^{2} (x) + cos^{2} (x))

sin^{3} (x) + cos^{2} (x) sin (x)

Apply exponent rule:

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

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

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

Factor out common term

sin (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Simplify

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

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

Use the basic trigonometric identity:

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

= tan (x)

We showed that the two sides could take the same form

\Rightarrow True

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

- tan (θ) cos (θ) = sin (- θ)

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

sin^{2} (2 x) = \frac{1}{2} (1 - cos (4 x))

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

Is sin(x)cos(x)+sin^3(x)sec(x)=tan(x) ?
The answer to whether sin(x)cos(x)+sin^3(x)sec(x)=tan(x) is True