Is 1/(sin(θ)cos(θ))-tan(θ)=cot(θ) ?

The answer to whether 1/(sin(θ)cos(θ))-tan(θ)=cot(θ) is True

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prove 1/(sin(θ)cos(θ))-tan(θ)=cot(θ)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

\frac{1}{cos ( θ ) sin ( θ )} - \frac{sin ( θ )}{cos ( θ )}

Least Common Multiplier of

cos (θ) sin (θ), cos (θ) : cos (θ) sin (θ)

cos (θ) sin (θ), cos (θ)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (θ) sin (θ)

cos (θ)

= cos (θ) sin (θ)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

cos (θ) sin (θ)

For

\frac{sin ( θ )}{cos ( θ )} :

multiply the denominator and numerator by

sin (θ)

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (θ)

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

Use the basic trigonometric identity:

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

= cot (θ)

We showed that the two sides could take the same form

\Rightarrow True

(- sin (x)) (3 cos (x)) = - 3 sin (x) cos (x)

cot (θ) \cdot sec (θ) = \frac{1}{cos ( θ )}

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

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

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

Is 1/(sin(θ)cos(θ))-tan(θ)=cot(θ) ?
The answer to whether 1/(sin(θ)cos(θ))-tan(θ)=cot(θ) is True