Question

demonstre: tg²a - sen²a = tg²a sen²a

Answer 1

Seja f(a) = tg²a - sen²a, g(a) = tg²a · sen²a, e h(a) = (sen²a - sen²a · cos²a)/cos²a

Para f(a)

$f(a)=\tan^2{a}-\sin^2{a}=\dfrac{\sin^2{a}}{\cos^2{a}}-\sin^2{a}=\dfrac{\sin^2{a}-\sin^2{a}\cdot\cos^2{a}}{\cos^2{a}}=h(a)$

Para g(a)

$g(a)=\tan^2{a}\cdot\sin^2{a}=\dfrac{\sin^2{a}}{\cos^2{a}}\cdot\sin^2{a}=\dfrac{\sin^2{a}}{\cos^2{a}}\cdot(1-\cos^2{a})\\\\=\dfrac{\sin^2{a}-\sin^2{a}\cos^2{a}}{\cos^2{a}}=h(a)$

Logo

$\left.\begin{matrix} {f\equiv h\\g\equiv h} \end{matrix}\right\}\Rightarrow~f\equiv g~\Leftrightarrow~\tan^2(a)-\sin^2{a}= \tan^2{a}\cdot\sin^2{a}$

Está demonstrado.