Is (cos(β))/(tan(β))=csc(β)-sin(β) ?

The answer to whether (cos(β))/(tan(β))=csc(β)-sin(β) is True

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prove (cos(β))/(tan(β))=csc(β)-sin(β)

prove

\frac{cos ( β )}{tan ( β )} = csc (β) - sin (β)

True

Solution steps

\frac{cos ( β )}{tan ( β )} = csc (β) - sin (β)

Manipulating left side

\frac{cos ( β )}{tan ( β )}

Express with sin, cos

\frac{cos ( β )}{tan ( β )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

cos (β) cos (β) = cos^{2} (β)

cos (β) cos (β)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (β) cos (β) = cos^{1 + 1} (β)

= cos^{1 + 1} (β)

Add the numbers:

1 + 1 = 2

= cos^{2} (β)

= \frac{cos ^{2} ( β )}{sin ( β )}

= \frac{cos ^{2} ( β )}{sin ( β )}

= \frac{cos ^{2} ( β )}{sin ( β )}

Manipulating right side

csc (β) - sin (β)

Express with sin, cos

csc (β) - sin (β)

Use the basic trigonometric identity:

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

= \frac{1}{sin ( β )} - sin (β)

Simplify

\frac{1}{sin ( β )} - sin (β) : \frac{1 - sin ^{2} ( β )}{sin ( β )}

\frac{1}{sin ( β )} - sin (β)

Convert element to fraction:

sin (β) = \frac{sin ( β ) sin ( β )}{sin ( β )}

= \frac{1}{sin ( β )} - \frac{sin ( β ) sin ( β )}{sin ( β )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 - sin ( β ) sin ( β )}{sin ( β )}

1 - sin (β) sin (β) = 1 - sin^{2} (β)

1 - sin (β) sin (β)

sin (β) sin (β) = sin^{2} (β)

sin (β) sin (β)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (β) sin (β) = sin^{1 + 1} (β)

= sin^{1 + 1} (β)

Add the numbers:

1 + 1 = 2

= sin^{2} (β)

= 1 - sin^{2} (β)

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

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

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

Rewrite using trig identities

\frac{1 - sin ^{2} ( β )}{sin ( β )}

Use the Pythagorean identity:

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

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

= \frac{cos ^{2} ( β )}{sin ( β )}

= \frac{cos ^{2} ( β )}{sin ( β )}

We showed that the two sides could take the same form

\Rightarrow True

cos (3 t) = 4 cos^{3} (t) - 3 cos (t)

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

sec (θ + π) = - sec (θ)

sec (2 x) = \frac{sec ^{2} ( x ) + sec ^{4} ( x )}{2 + sec ^{2} ( x ) - sec ^{4} ( x )}

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

Is (cos(β))/(tan(β))=csc(β)-sin(β) ?
The answer to whether (cos(β))/(tan(β))=csc(β)-sin(β) is True