Is tan(b)cos(b)+cot(b)cos(b)=csc(b) ?

The answer to whether tan(b)cos(b)+cot(b)cos(b)=csc(b) is True

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Popular >

prove tan(b)cos(b)+cot(b)cos(b)=csc(b)

Solution

prove

tan (b) cos (b) + cot (b) cos (b) = csc (b)

Solution

True

Solution steps

tan (b) cos (b) + cot (b) cos (b) = csc (b)

Manipulating left side

tan (b) cos (b) + cot (b) cos (b)

Express with sin, cos

cos (b) cot (b) + cos (b) tan (b)

Use the basic trigonometric identity:

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

= cos (b) \frac{cos ( b )}{sin ( b )} + cos (b) tan (b)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (b)

= sin (b)

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

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

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

Popular Examples

prove 1+cot^2(z)=csc^2(z)prove

1 + cot^{2} (z) = csc^{2} (z)

prove csc(x)-1=(2sin(x))/(sin(x))prove

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

prove 2sin^2(θ)-cos(2θ)=1 prove

2 sin^{2} (θ) - cos (2 θ) = 1

prove 1/(csc(+1))+1/(sin(x)+1)=1 prove

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

prove ((tan^2(t)))/(sec(t))=sec(t)-cos(t)prove

\frac{( tan ^{2} ( t ) )}{sec ( t )} = sec (t) - cos (t)

Frequently Asked Questions (FAQ)

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