Is (csc(B))/(tan(B)+cot(B))=cos(B) ?

The answer to whether (csc(B))/(tan(B)+cot(B))=cos(B) is True

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prove (csc(B))/(tan(B)+cot(B))=cos(B)

prove

\frac{csc ( B )}{tan ( B ) + cot ( B )} = cos (B)

True

Solution steps

\frac{csc ( B )}{tan ( B ) + cot ( B )} = cos (B)

Manipulating left side

\frac{csc ( B )}{tan ( B ) + cot ( B )}

Express with sin, cos

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

Rewrite using trig identities

= cos (B)

We showed that the two sides could take the same form

\Rightarrow True

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

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

sin (\frac{17 π}{2} + x) = cos (x)

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

tan^{2} (x) + \frac{1}{sin ( x ) csc ( x )} = sec^{2} (x)

Is (csc(B))/(tan(B)+cot(B))=cos(B) ?
The answer to whether (csc(B))/(tan(B)+cot(B))=cos(B) is True