Is (cos(3a)cos(2a))/(sin(3a)sin(2a))=(cot(2a))/(tan(3a)) ?

The answer to whether (cos(3a)cos(2a))/(sin(3a)sin(2a))=(cot(2a))/(tan(3a)) is True

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prove (cos(3a)cos(2a))/(sin(3a)sin(2a))=(cot(2a))/(tan(3a))

prove

\frac{cos ( 3 a ) cos ( 2 a )}{sin ( 3 a ) sin ( 2 a )} = \frac{cot ( 2 a )}{tan ( 3 a )}

True

Solution steps

\frac{cos ( 3 a ) cos ( 2 a )}{sin ( 3 a ) sin ( 2 a )} = \frac{cot ( 2 a )}{tan ( 3 a )}

Manipulating right side

\frac{cot ( 2 a )}{tan ( 3 a )}

Express with sin, cos

\frac{cot ( 2 a )}{tan ( 3 a )}

Use the basic trigonometric identity:

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

= \frac{\frac{c o s ( 2 a )}{s i n ( 2 a )}}{tan ( 3 a )}

Use the basic trigonometric identity:

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

= \frac{\frac{c o s ( 2 a )}{s i n ( 2 a )}}{\frac{s i n ( 3 a )}{c o s ( 3 a )}}

Simplify

\frac{\frac{c o s ( 2 a )}{s i n ( 2 a )}}{\frac{s i n ( 3 a )}{c o s ( 3 a )}} : \frac{cos ( 2 a ) cos ( 3 a )}{sin ( 2 a ) sin ( 3 a )}

\frac{\frac{c o s ( 2 a )}{s i n ( 2 a )}}{\frac{s i n ( 3 a )}{c o s ( 3 a )}}

Divide fractions:

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

= \frac{cos ( 2 a ) cos ( 3 a )}{sin ( 2 a ) sin ( 3 a )}

= \frac{cos ( 2 a ) cos ( 3 a )}{sin ( 2 a ) sin ( 3 a )}

= \frac{cos ( 2 a ) cos ( 3 a )}{sin ( 2 a ) sin ( 3 a )}

= \frac{cos ( 3 a ) cos ( 2 a )}{sin ( 3 a ) sin ( 2 a )}

We showed that the two sides could take the same form

\Rightarrow True

\frac{1}{cos ^{2} ( θ )} = tan^{2} (θ) + 1

(sin (θ) - 1) (tan (θ) + sec (θ)) = - cos (θ)

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

sin (b) cos^{2} (b) - sin (b) = - sin^{3} (b)

\frac{1}{2} (cot (θ) - tan (θ)) = cot (2 θ)

Is (cos(3a)cos(2a))/(sin(3a)sin(2a))=(cot(2a))/(tan(3a)) ?
The answer to whether (cos(3a)cos(2a))/(sin(3a)sin(2a))=(cot(2a))/(tan(3a)) is True