Is tan(θ)+cot(θ)=sec(θ)*csc(θ) ?

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

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prove tan(θ)+cot(θ)=sec(θ)*csc(θ)

prove

tan (θ) + cot (θ) = sec (θ) \cdot csc (θ)

True

Solution steps

tan (θ) + cot (θ) = sec (θ) csc (θ)

Manipulating left side

tan (θ) + cot (θ)

Express with sin, cos

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

Rewrite using trig identities

= \frac{1}{cos ( θ ) sin ( θ )}

Rewrite using trig identities

sec (θ) csc (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{cos ( x )}{csc ( x ) + 1} + \frac{cos ( x )}{csc ( x ) - 1} = 2 tan (x)

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

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

cot (\frac{π}{2} - u) = tan (u)

\frac{sec ^{4} ( x ) - 1}{tan ^{2} ( x )} = tan^{2} (x) + 2

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