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

sin^{2} (θ) (csc^{2} (θ) - 1) = cos^{2} (θ)

cot (α) sec (α) = csc (α)

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

\frac{1}{sec ( a ) + tan ( a )} = sec (a) - tan (a)

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

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