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

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

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

prove

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

True

Solution steps

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

Manipulating left side

sin (θ) cos (θ) + sin^{3} (θ) sec (θ)

Express with sin, cos

cos (θ) sin (θ) + sec (θ) sin^{3} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

\frac{1}{cos ( θ )} sin^{3} (θ)

Multiply fractions:

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

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

Multiply:

1 \cdot sin^{3} (θ) = sin^{3} (θ)

= \frac{sin ^{3} ( θ )}{cos ( θ )}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

cos (θ) sin (θ) cos (θ) + sin^{3} (θ)

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

cos (θ) sin (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= sin (θ) cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin (θ) cos^{2} (θ)

= cos^{2} (θ) sin (θ) + sin^{3} (θ)

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

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

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

Factor

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

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

Factor

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

sin^{3} (θ) + cos^{2} (θ) sin (θ)

Apply exponent rule:

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

sin^{3} (θ) = sin (θ) sin^{2} (θ)

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

Factor out common term

sin (θ)

= sin (θ) (sin^{2} (θ) + cos^{2} (θ))

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= \frac{1 \cdot sin ( θ )}{cos ( θ )}

Simplify

= \frac{sin ( θ )}{cos ( θ )}

= \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

= tan (θ)

We showed that the two sides could take the same form

\Rightarrow True

tan (A) sin (A) = sec (A) - cos (A)

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

sin (x) sec (x) = 1

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

tan (a - b) = sin (a - b) sec (a - b)

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