Is (1+sec(θ))/(csc(θ))-tan(θ)=sin(θ) ?

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

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prove (1+sec(θ))/(csc(θ))-tan(θ)=sin(θ)

prove

\frac{1 + sec ( θ )}{csc ( θ )} - tan (θ) = sin (θ)

True

Solution steps

\frac{1 + sec ( θ )}{csc ( θ )} - tan (θ) = sin (θ)

Manipulating left side

\frac{1 + sec ( θ )}{csc ( θ )} - tan (θ)

Express with sin, cos

= sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{1 + cot ( x )}{1 + tan ( x )} = cot (x)

cos (α + β) + cos (α - β) = 2 cos (α) cos (β)

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

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

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

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