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

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

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

prove

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

True

Solution steps

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

= cot (A) cos (A)

We showed that the two sides could take the same form

\Rightarrow True

cos (x) sec (x) + tan (x) = 1 + sin (x)

\frac{cos ( θ )}{sec ( θ ) - tan ( θ )} = 1 + sin (x)

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

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

1 - sin^{5} (x) = cos^{2} (x)

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