Is tan(β)=(cos(β))/(sin(β)cot^2(β)) ?

The answer to whether tan(β)=(cos(β))/(sin(β)cot^2(β)) is True

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prove tan(β)=(cos(β))/(sin(β)cot^2(β))

prove

tan (β) = \frac{cos ( β )}{sin ( β ) cot ^{2} ( β )}

True

Solution steps

tan (β) = \frac{cos ( β )}{sin ( β ) cot ^{2} ( β )}

Manipulating right side

\frac{cos ( β )}{sin ( β ) cot ^{2} ( β )}

Express with sin, cos

\frac{cos ( β )}{cot ^{2} ( β ) sin ( β )}

Use the basic trigonometric identity:

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

= \frac{cos ( β )}{( \frac{c o s ( β )}{s i n ( β )} ) ^{2} sin ( β )}

Simplify

\frac{cos ( β )}{( \frac{c o s ( β )}{s i n ( β )} ) ^{2} sin ( β )} : \frac{sin ( β )}{cos ( β )}

\frac{cos ( β )}{( \frac{c o s ( β )}{s i n ( β )} ) ^{2} sin ( β )}

(\frac{cos ( β )}{sin ( β )})^{2} = \frac{cos ^{2} ( β )}{sin ^{2} ( β )}

(\frac{cos ( β )}{sin ( β )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{cos ^{2} ( β )}{sin ^{2} ( β )}

= \frac{cos ( β )}{\frac{c o s ^{2} ( β )}{s i n ^{2} ( β )} sin ( β )}

Multiply

\frac{cos ^{2} ( β )}{sin ^{2} ( β )} sin (β) : \frac{cos ^{2} ( β )}{sin ( β )}

\frac{cos ^{2} ( β )}{sin ^{2} ( β )} sin (β)

Multiply fractions:

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

= \frac{cos ^{2} ( β ) sin ( β )}{sin ^{2} ( β )}

Cancel the common factor:

sin (β)

= \frac{cos ^{2} ( β )}{sin ( β )}

= \frac{cos ( β )}{\frac{c o s ^{2} ( β )}{s i n ( β )}}

Apply the fraction rule:

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

= \frac{cos ( β ) sin ( β )}{cos ^{2} ( β )}

Cancel the common factor:

cos (β)

= \frac{sin ( β )}{cos ( β )}

= \frac{sin ( β )}{cos ( β )}

= \frac{sin ( β )}{cos ( β )}

Use the basic trigonometric identity:

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

= tan (β)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{sin ^{2} ( x )}{cos ( x )} = sin (x) tan (x)

p ro v e csc^{2} (x) sec^{2} (x) = sec^{2} (x) + csc^{2} (x)

p ro v e \frac{sin ( x ) + cos ( x )}{cos ( x )} = 1 + tan (x)

p ro v e \frac{1 + sin ( - x )}{cos ( x ) tan ( x ) - 1} = - 1

p ro v e sec (A) - \frac{cos ( A )}{1 + sin ( A )} = tan (A)

Is tan(β)=(cos(β))/(sin(β)cot^2(β)) ?
The answer to whether tan(β)=(cos(β))/(sin(β)cot^2(β)) is True