Is sin(x)tan(x)-csc(x)tan(x)=-cot(x) ?

The answer to whether sin(x)tan(x)-csc(x)tan(x)=-cot(x) is False

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prove sin(x)tan(x)-csc(x)tan(x)=-cot(x)

prove

sin (x) tan (x) - csc (x) tan (x) = - cot (x)

False

Solution steps

sin (x) tan (x) - csc (x) tan (x) = - cot (x)

Show that the two sides are not equal

For

sin (x) tan (x) - csc (x) tan (x) = - cot (x)

plug in

x = 1

sin (1) tan (1) - csc (1) tan (1) = - 0.54030 \dots

sin (1) tan (1) - csc (1) tan (1)

Refine to a decimal form

= - 0.54030 \dots

- cot (1) = - 0.64209 \dots

- cot (1)

Refine to a decimal form

= - 0.64209 \dots

The two sides are not equal

\Rightarrow False

sin (2 x) csc (2 x) = 1

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

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

\frac{( cos ( π + x ) )}{cos ( \frac{3 π}{2} - x )} = cot (x)

1 + (sec^{2} (x)) (sin^{2} (x)) = sec^{2} (x)

Is sin(x)tan(x)-csc(x)tan(x)=-cot(x) ?
The answer to whether sin(x)tan(x)-csc(x)tan(x)=-cot(x) is False