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

The answer to whether (csc(θ)+sin(θ))/(tan(θ)+1)=csc(θ) is False

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

\Rightarrow False

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

sin (3 θ) = 3 cos^{2} (θ) sin (θ) - sin^{3} (θ)

\frac{tan ( θ ) cos ( θ )}{sin ( θ )} = 1

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

sec (x) - cos (x) = sin (x) (tan (x))

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