Is (tan^2(x))/(sin^2(x))-1=tan^2(x) ?

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Rewrite using trig identities

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

Rewrite using trig identities

tan^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{1}{tan ( x )} = cot (x)

\frac{1 + tan ( θ )}{cos ( θ ) + sin ( θ )} = sec (θ)

\frac{cos ( x )}{sec ( x ) + tan ( x )} = 1 - sin (x)

tan (x) \cdot sin (x) + cos (x) = sec (x)

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