Is 6sin^2(x)-3cos^2(x)=9sin^2(x)-3 ?

The answer to whether 6sin^2(x)-3cos^2(x)=9sin^2(x)-3 is True

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prove 6sin^2(x)-3cos^2(x)=9sin^2(x)-3

prove

6 sin^{2} (x) - 3 cos^{2} (x) = 9 sin^{2} (x) - 3

True

Solution steps

6 sin^{2} (x) - 3 cos^{2} (x) = 9 sin^{2} (x) - 3

Manipulating left side

6 sin^{2} (x) - 3 cos^{2} (x)

Rewrite using trig identities

= 9 sin^{2} (x) - 3

We showed that the two sides could take the same form

\Rightarrow True

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

tan^{2} (θ) + cot^{2} (θ) = sec^{2} (θ) csc^{2} (θ)

cos^{2} (u) = 1 - sin^{2} (u)

sin^{2} (csc^{2} (x) - 1) = cos^{2} (x)

sin^{2} (x) - sin^{4} (x) = 1 - 2 sin^{2} (x)

Is 6sin^2(x)-3cos^2(x)=9sin^2(x)-3 ?
The answer to whether 6sin^2(x)-3cos^2(x)=9sin^2(x)-3 is True