Is cot(θ)sec^2(θ)-cot(θ)=tan(θ) ?

The answer to whether cot(θ)sec^2(θ)-cot(θ)=tan(θ) is True

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prove cot(θ)sec^2(θ)-cot(θ)=tan(θ)

prove

cot (θ) sec^{2} (θ) - cot (θ) = tan (θ)

True

Solution steps

cot (θ) sec^{2} (θ) - cot (θ) = tan (θ)

Manipulating left side

cot (θ) sec^{2} (θ) - cot (θ)

Express with sin, cos

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

Rewrite using trig identities

= \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

2 cos (x - \frac{π}{4}) = cos (x) + sin (x)

(cos^{2} (x) - 1) (tan^{2} (x) + 1) = - tan^{2} (x)

7 csc^{2} (θ) - 5 cot^{2} (θ) = 2 csc^{2} (θ) + 5

cos (θ) = \frac{cot ( θ )}{csc ( θ )}

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

Is cot(θ)sec^2(θ)-cot(θ)=tan(θ) ?
The answer to whether cot(θ)sec^2(θ)-cot(θ)=tan(θ) is True