Is cot(u)-csc(u)=(sin(u))/(1+cos(u)) ?

The answer to whether cot(u)-csc(u)=(sin(u))/(1+cos(u)) is False

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prove cot(u)-csc(u)=(sin(u))/(1+cos(u))

prove

cot (u) - csc (u) = \frac{sin ( u )}{1 + cos ( u )}

False

Solution steps

cot (u) - csc (u) = \frac{sin ( u )}{1 + cos ( u )}

Show that the two sides are not equal

For

cot (u) - csc (u) = \frac{sin ( u )}{1 + cos ( u )}

plug in

u = 1

cot (1) - csc (1) = - 0.54630 \dots

cot (1) - csc (1)

Refine to a decimal form

= - 0.54630 \dots

\frac{sin ( 1 )}{1 + cos ( 1 )} = 0.54630 \dots

\frac{sin ( 1 )}{1 + cos ( 1 )}

Refine to a decimal form

= 0.54630 \dots

The two sides are not equal

\Rightarrow False

cos (θ) tan (- θ) = - sin (θ)

tan^{3} (x) cot (x) = sec^{2} (x) - 1

\frac{sec ^{2} ( u ) - 1}{sec ^{2} ( u )} = sin^{2} (u)

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

sin (\frac{7 π}{6}) = - \frac{1}{2}

Is cot(u)-csc(u)=(sin(u))/(1+cos(u)) ?
The answer to whether cot(u)-csc(u)=(sin(u))/(1+cos(u)) is False