Is (tan^2(θ))/(sin^2(θ))-1=tan^2(θ) ?

The answer to whether (tan^2(θ))/(sin^2(θ))-1=tan^2(θ) is True

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prove (tan^2(θ))/(sin^2(θ))-1=tan^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

\frac{tan ^{2} ( θ )}{sin ^{2} ( θ )} - 1

Express with sin, cos

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

Rewrite using trig identities

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

Rewrite using trig identities

tan^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

sec (θ + \frac{π}{2}) = csc (θ)

\frac{1 + csc ( θ )}{sec ( θ )} - cot (θ) = cos (θ)

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

cos (A) csc (A) = cot (A)

Is (tan^2(θ))/(sin^2(θ))-1=tan^2(θ) ?
The answer to whether (tan^2(θ))/(sin^2(θ))-1=tan^2(θ) is True