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

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

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

prove

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

True

Solution steps

cos (a) cot (a) = csc (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 )}

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

\frac{1 + tan ^{2} ( x )}{cot ^{2} ( x ) - 1} = tan^{2} (x)

cos (2 A) = 2 cos^{2} (A) - 2

(sec^{2} (x) - 1) cot^{2} (x) = 1

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

\frac{1 + cos ( θ )}{tan ( θ ) + sin ( θ )} = cot (θ)

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