Is sin^2(θ)+sin^2(θ)tan^2(θ)=tan^2(θ) ?

The answer to whether sin^2(θ)+sin^2(θ)tan^2(θ)=tan^2(θ) is True

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prove sin^2(θ)+sin^2(θ)tan^2(θ)=tan^2(θ)

prove

sin^{2} (θ) + sin^{2} (θ) tan^{2} (θ) = tan^{2} (θ)

True

Solution steps

sin^{2} (θ) + sin^{2} (θ) tan^{2} (θ) = tan^{2} (θ)

Manipulating left side

sin^{2} (θ) + sin^{2} (θ) tan^{2} (θ)

Factor

sin^{2} (θ) + sin^{2} (θ) tan^{2} (θ) : sin^{2} (θ) (1 + tan^{2} (θ))

= (1 + tan^{2} (θ)) sin^{2} (θ)

Rewrite using trig identities

= sec^{2} (θ) sin^{2} (θ)

Express with sin, cos

= \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

1 + sin (2 θ) = (cos (θ) sin (θ))^{2}

sec (θ) csc (θ) = tan (θ) + cot (θ)

2 tan (0^{\circ}) = sin (0^{\circ})

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

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

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