Is (1+tan^2(θ))/(csc(θ))=tan(θ)sec(θ) ?

The answer to whether (1+tan^2(θ))/(csc(θ))=tan(θ)sec(θ) is True

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prove (1+tan^2(θ))/(csc(θ))=tan(θ)sec(θ)

prove

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

True

Solution steps

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

Manipulating left side

\frac{1 + tan ^{2} ( θ )}{csc ( θ )}

Express with sin, cos

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

Rewrite using trig identities

= \frac{sin ( θ )}{cos ^{2} ( θ )}

Manipulating right side

tan (θ) sec (θ)

Express with sin, cos

= \frac{sin ( θ )}{cos ^{2} ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

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

4 sin^{2} (x) + 4 cos^{2} (x) = 4

tan^{2} (a) = (tan (a))^{2}

(1 - cos (θ^{2})) (1 + cot (θ^{2})) = 1

cos (2 x) = \frac{2 - sec ^{2} ( x )}{sec ^{2} ( x )}

Is (1+tan^2(θ))/(csc(θ))=tan(θ)sec(θ) ?
The answer to whether (1+tan^2(θ))/(csc(θ))=tan(θ)sec(θ) is True