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

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

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

prove

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

True

Solution steps

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

Manipulating left 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 )}

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

cos (a) cot (a)

cos (a) cot (a)

We showed that the two sides could take the same form

\Rightarrow True

cos^{2} (α) + cos^{2} (α) tan^{2} (α) = 1

cos (\frac{π}{7}) = sin (\frac{π}{2} - \frac{π}{7})

sin (\frac{π}{4} + x) = 2 (cos (x) + sin (x))

sec (x) csc (x) = cot (x) + tan (x)

arctan (x) = \frac{arcsin ( x )}{arccos ( x )}

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