Is (csc(θ)-cot(θ))(1+sec(θ))=tan(θ) ?

The answer to whether (csc(θ)-cot(θ))(1+sec(θ))=tan(θ) is True

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prove (csc(θ)-cot(θ))(1+sec(θ))=tan(θ)

prove

(csc (θ) - cot (θ)) (1 + sec (θ)) = tan (θ)

True

Solution steps

(csc (θ) - cot (θ)) (1 + sec (θ)) = tan (θ)

Manipulating left side

(csc (θ) - cot (θ)) (1 + sec (θ))

Express with sin, cos

= \frac{( 1 + cos ( θ ) ) ( 1 - cos ( 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

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

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

cos (3 θ) = cos (θ)

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

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

Is (csc(θ)-cot(θ))(1+sec(θ))=tan(θ) ?
The answer to whether (csc(θ)-cot(θ))(1+sec(θ))=tan(θ) is True