Is sin(B)+cos(B)cot(B)=csc(B) ?

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

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prove sin(B)+cos(B)cot(B)=csc(B)

prove

sin (B) + cos (B) cot (B) = csc (B)

True

Solution steps

sin (B) + cos (B) cot (B) = csc (B)

Manipulating left side

sin (B) + cos (B) cot (B)

Express with sin, cos

sin (B) + cos (B) cot (B)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

cos (B) \frac{cos ( B )}{sin ( B )}

Multiply fractions:

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

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

cos (B) cos (B) = cos^{2} (B)

cos (B) cos (B)

Apply exponent rule:

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

cos (B) cos (B) = cos^{1 + 1} (B)

= cos^{1 + 1} (B)

Add the numbers:

1 + 1 = 2

= cos^{2} (B)

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

sin (B) sin (B) + cos^{2} (B)

sin (B) sin (B) = sin^{2} (B)

sin (B) sin (B)

Apply exponent rule:

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

sin (B) sin (B) = sin^{1 + 1} (B)

= sin^{1 + 1} (B)

Add the numbers:

1 + 1 = 2

= sin^{2} (B)

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= \frac{1}{sin ( B )}

= \frac{1}{sin ( B )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( B )}}

Simplify

\frac{1}{\frac{1}{c s c ( B )}}

Apply the fraction rule:

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

= \frac{csc ( B )}{1}

Apply rule

\frac{a}{1} = a

= csc (B)

csc (B)

csc (B)

We showed that the two sides could take the same form

\Rightarrow True

3 sin (5 π - x) = 3 sin (x)

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

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

\frac{cos ^{2} ( α )}{1 - sin ( α )} = 1 + sin (α)

2 sin^{2} (θ) - 1 = 1 - 2 cos^{2} (θ)

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