Is cot(θ)=sec(θ)csc(θ)-tan(θ) ?

The answer to whether cot(θ)=sec(θ)csc(θ)-tan(θ) is True

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prove cot(θ)=sec(θ)csc(θ)-tan(θ)

prove

cot (θ) = sec (θ) csc (θ) - tan (θ)

True

Solution steps

cot (θ) = sec (θ) csc (θ) - tan (θ)

Manipulating right side

sec (θ) csc (θ) - tan (θ)

Express with sin, cos

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

Rewrite using trig identities

= \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^{2} (β) csc^{2} (β) - sin^{2} (β) = cos^{2} (β)

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

cos (x) = sin (x)

csc (x) - cot (x) = tan (\frac{x}{2})

cot^{2} (a) - cos^{2} (a) = cot^{2} (a) cos^{2} (a)

Is cot(θ)=sec(θ)csc(θ)-tan(θ) ?
The answer to whether cot(θ)=sec(θ)csc(θ)-tan(θ) is True