Is sin(β)+cos(β)cot(β)=csc(β) ?

The answer to whether sin(β)+cos(β)cot(β)=csc(β) is True

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prove sin(β)+cos(β)cot(β)=csc(β)

prove

sin (β) + cos (β) cot (β) = csc (β)

True

Solution steps

sin (β) + cos (β) cot (β) = csc (β)

Manipulating left side

sin (β) + cos (β) cot (β)

Express with sin, cos

sin (β) + cos (β) cot (β)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

cos (β) \frac{cos ( β )}{sin ( β )}

Multiply fractions:

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

= \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 )}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

sin (β) sin (β) + cos^{2} (β) = sin^{2} (β) + cos^{2} (β)

sin (β) sin (β) + cos^{2} (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)

= sin^{2} (β) + cos^{2} (β)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= \frac{1}{sin ( β )}

= \frac{1}{sin ( β )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( β )}}

Simplify

\frac{1}{\frac{1}{c s c ( β )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{csc ( β )}{1}

Apply rule

\frac{a}{1} = a

= csc (β)

csc (β)

csc (β)

We showed that the two sides could take the same form

\Rightarrow True

cos (\frac{3 π}{2} + x) = - cos (x)

sec (x) + tan (x) = \frac{1}{sec ( x ) + tan ( x )}

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

sin (θ) cos (θ) = tan (θ)

sin (θ) cos (θ) = cot (θ)

Is sin(β)+cos(β)cot(β)=csc(β) ?
The answer to whether sin(β)+cos(β)cot(β)=csc(β) is True