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

The answer to whether (csc(x)sin(x))/(tan(x))=cot(x) is True

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

prove

\frac{csc ( x ) sin ( x )}{tan ( x )} = cot (x)

True

Solution steps

\frac{csc ( x ) sin ( x )}{tan ( x )} = cot (x)

Manipulating left side

\frac{csc ( x ) sin ( x )}{tan ( x )}

Express with sin, cos

\frac{csc ( x ) sin ( x )}{tan ( x )}

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= \frac{\frac{1}{s i n ( x )} sin ( x )}{tan ( x )}

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{\frac{1}{s i n ( x )} sin ( x )}{\frac{s i n ( x )}{c o s ( x )}}

Simplify

\frac{\frac{1}{s i n ( x )} sin ( x )}{\frac{s i n ( x )}{c o s ( x )}} : \frac{cos ( x )}{sin ( x )}

\frac{\frac{1}{s i n ( x )} sin ( x )}{\frac{s i n ( x )}{c o s ( x )}}

Apply the fraction rule:

\frac{a}{\frac{b}{c}} = \frac{a \cdot c}{b}

= \frac{\frac{1}{s i n ( x )} sin ( x ) cos ( x )}{sin ( x )}

Multiply

\frac{1}{sin ( x )} sin (x) cos (x) : cos (x)

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Cancel the common factor:

sin (x)

= 1 \cdot cos (x)

Multiply:

1 \cdot cos (x) = cos (x)

= cos (x)

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

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

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

Use the basic trigonometric identity:

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

= cot (x)

We showed that the two sides could take the same form

\Rightarrow True

cos (- 2 x) = cosh^{2} (x) + sinh^{2} (x)

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

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

sec (2 x) + tan (2 x) = \frac{( cos ( x ) + sin ( x ) )}{( cos ( x ) - sin ( x ) )}

\frac{sec ( - 1 )}{1 - cos ( x )} = sec (x)

Is (csc(x)sin(x))/(tan(x))=cot(x) ?
The answer to whether (csc(x)sin(x))/(tan(x))=cot(x) is True