Is sin(θ)+cos(θ)cot(θ)=csc(θ) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove sin(θ)+cos(θ)cot(θ)=csc(θ)

prove

sin (θ) + cos (θ) cot (θ) = csc (θ)

True

Solution steps

sin (θ) + cos (θ) cot (θ) = csc (θ)

Manipulating left side

sin (θ) + cos (θ) cot (θ)

Express with sin, cos

sin (θ) + cos (θ) cot (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

cos (θ) \frac{cos ( θ )}{sin ( θ )}

Multiply fractions:

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

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

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

= \frac{cos ^{2} ( θ )}{sin ( θ )}

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

sin (θ) sin (θ) + cos^{2} (θ)

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= sin^{2} (θ) + cos^{2} (θ)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( θ )}}

Simplify

\frac{1}{\frac{1}{c s c ( θ )}}

Apply the fraction rule:

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

= \frac{csc ( θ )}{1}

Apply rule

\frac{a}{1} = a

= csc (θ)

csc (θ)

csc (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

sin (x - \frac{π}{6}) + cos (x) = sin (x + \frac{π}{6})

sec (θ) sin (θ) = tan (θ)

(1 + tan (x))^{2} = sec^{2} (x) + 2 tan (x)

\frac{sec ^{2} ( x )}{2 tan ( x )} = csc (2 x)

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