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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

prove

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

True

Solution steps

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

Manipulating left side

csc (B) - sin (B)

Express with sin, cos

csc (B) - sin (B)

Use the basic trigonometric identity:

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

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

Simplify

\frac{1}{sin ( B )} - sin (B) : \frac{1 - sin ^{2} ( B )}{sin ( B )}

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

Convert element to fraction:

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

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

1 - sin (B) sin (B) = 1 - sin^{2} (B)

1 - 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)

= 1 - sin^{2} (B)

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

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

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

Rewrite using trig identities

\frac{1 - sin ^{2} ( B )}{sin ( B )}

Use the Pythagorean identity:

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

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

cos (B) cot (B)

cos (B) cot (B)

= cot (B) cos (B)

We showed that the two sides could take the same form

\Rightarrow True

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

cos (2 a + a) = - 3 cos (a) + 4 cos^{3} (a)

cot (2 θ) = \frac{1}{2} sec (θ) csc (θ) - tan (θ)

csc^{2} (θ) - cos (θ) sec (θ) = cot^{2} (θ)

sin (x) + cos (x) = 1

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