Is csc(2A)+cot(2A)=cot(A) ?

The answer to whether csc(2A)+cot(2A)=cot(A) is True

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prove csc(2A)+cot(2A)=cot(A)

prove

csc (2 A) + cot (2 A) = cot (A)

True

Solution steps

csc (2 A) + cot (2 A) = cot (A)

Manipulating left side

csc (2 A) + cot (2 A)

Express with sin, cos

cot (2 A) + csc (2 A)

Use the basic trigonometric identity:

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

= \frac{cos ( 2 A )}{sin ( 2 A )} + csc (2 A)

Use the basic trigonometric identity:

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

= \frac{cos ( 2 A )}{sin ( 2 A )} + \frac{1}{sin ( 2 A )}

Simplify

\frac{cos ( 2 A )}{sin ( 2 A )} + \frac{1}{sin ( 2 A )} : \frac{cos ( 2 A ) + 1}{sin ( 2 A )}

\frac{cos ( 2 A )}{sin ( 2 A )} + \frac{1}{sin ( 2 A )}

Apply rule

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

= \frac{cos ( 2 A ) + 1}{sin ( 2 A )}

= \frac{cos ( 2 A ) + 1}{sin ( 2 A )}

= \frac{1 + cos ( 2 A )}{sin ( 2 A )}

Rewrite using trig identities

\frac{1 + cos ( 2 A )}{sin ( 2 A )}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= \frac{1 + cos ( 2 A )}{2 sin ( A ) cos ( A )}

Use the Double Angle identity:

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

= \frac{1 + 2 cos ^{2} ( A ) - 1}{2 sin ( A ) cos ( A )}

Simplify

\frac{1 + 2 cos ^{2} ( A ) - 1}{2 sin ( A ) cos ( A )} : \frac{cos ( A )}{sin ( A )}

\frac{1 + 2 cos ^{2} ( A ) - 1}{2 sin ( A ) cos ( A )}

1 + 2 cos^{2} (A) - 1 = 2 cos^{2} (A)

1 + 2 cos^{2} (A) - 1

Group like terms

= 2 cos^{2} (A) + 1 - 1

1 - 1 = 0

= 2 cos^{2} (A)

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

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

cos (A)

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

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

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

Use the basic trigonometric identity:

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

= cot (A)

We showed that the two sides could take the same form

\Rightarrow True

Is csc(2A)+cot(2A)=cot(A) ?
The answer to whether csc(2A)+cot(2A)=cot(A) is True