Is cos(θ)cot(θ)+sin^2(θ)csc(θ)=csc(θ) ?

The answer to whether cos(θ)cot(θ)+sin^2(θ)csc(θ)=csc(θ) is True

Popular >

prove cos(θ)cot(θ)+sin^2(θ)csc(θ)=csc(θ)

prove

cos (θ) cot (θ) + sin^{2} (θ) csc (θ) = csc (θ)

True

Solution steps

cos (θ) cot (θ) + sin^{2} (θ) csc (θ) = csc (θ)

Manipulating left side

cos (θ) cot (θ) + sin^{2} (θ) csc (θ)

Express with sin, cos

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

Rewrite using trig identities

= \frac{1}{sin ( θ )}

Rewrite using trig identities

csc (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

cos (a + b) - cos (a - b) = - 2 sin (a) sin (b)

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

sin (α + β) = sin (α) cos (β) + sin (β) cos (α)

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

Is cos(θ)cot(θ)+sin^2(θ)csc(θ)=csc(θ) ?
The answer to whether cos(θ)cot(θ)+sin^2(θ)csc(θ)=csc(θ) is True