Is (csc(θ)+sec(θ))/(tan(+1))=csc(θ) ?

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

θ = 1

\frac{csc ( 1 ) + sec ( 1 )}{tan ( 1 )} = 1.95145 \dots

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

Refine to a decimal form

= 1.95145 \dots

csc (1) = 1.18839 \dots

csc (1)

Refine to a decimal form

= 1.18839 \dots

The two sides are not equal

\Rightarrow False

\frac{csc ( x )}{csc ( 2 x )} = 2 cos (x)

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

\frac{1}{cot ( x )} = \frac{sin ( x )}{cos ( x )}

cot (x) - 1 = cos (x) (csc (x) - sec (x))

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

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