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

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

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

prove

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

True

Solution steps

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

Manipulating left side

csc (B) - sin (B)

Express with sin, cos

= \frac{1 - sin ^{2} ( B )}{sin ( B )}

Rewrite using trig identities

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

Rewrite using trig identities

cos (B) cot (B)

= cot (B) cos (B)

We showed that the two sides could take the same form

\Rightarrow True

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

cos (2 a + a) = - 3 cos (a) + 4 cos^{3} (a)

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

csc^{2} (θ) - cos (θ) sec (θ) = cot^{2} (θ)

sin (x) + cos (x) = 1

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