Is (cos(a))/(tan(a))=csc(a)-sin(a) ?

The answer to whether (cos(a))/(tan(a))=csc(a)-sin(a) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove (cos(a))/(tan(a))=csc(a)-sin(a)

prove

\frac{cos ( a )}{tan ( a )} = csc (a) - sin (a)

True

Solution steps

\frac{cos ( a )}{tan ( a )} = csc (a) - sin (a)

Manipulating left side

\frac{cos ( a )}{tan ( a )}

Express with sin, cos

\frac{cos ( a )}{tan ( a )}

Use the basic trigonometric identity:

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

= \frac{cos ( a )}{\frac{s i n ( a )}{c o s ( a )}}

Simplify

\frac{cos ( a )}{\frac{s i n ( a )}{c o s ( a )}} : \frac{cos ^{2} ( a )}{sin ( a )}

\frac{cos ( a )}{\frac{s i n ( a )}{c o s ( a )}}

Apply the fraction rule:

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

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

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

cos (a) cos (a)

Apply exponent rule:

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

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

= cos^{1 + 1} (a)

Add the numbers:

1 + 1 = 2

= cos^{2} (a)

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

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

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

Manipulating right side

csc (a) - sin (a)

Express with sin, cos

csc (a) - sin (a)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

1 - sin (a) sin (a)

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

sin (a) sin (a)

Apply exponent rule:

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

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

= sin^{1 + 1} (a)

Add the numbers:

1 + 1 = 2

= sin^{2} (a)

= 1 - sin^{2} (a)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

cot (t) + \frac{sin ( t )}{1 + cos ( t )} = csc (t)

sin (π + θ) = - sin (θ)

\frac{tan ( x ) cot ( x )}{sin ( x )} = csc (x)

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

Is (cos(a))/(tan(a))=csc(a)-sin(a) ?
The answer to whether (cos(a))/(tan(a))=csc(a)-sin(a) is True