Question

Demonstre as seguintes identidades: sec^2 x + cossec^2 x = sec^2 x . cossec^2 x

Answer 1

$\mathsf{sec^{n}x=\dfrac{1}{cos^{n}x}=\bigg(\dfrac{1}{cos\,x}\bigg)^{n}~~~~~~~~~~cossec^{n}x=\dfrac{1}{sen^{n}x}=\bigg(\dfrac{1}{sen\,x}\bigg)^{n}}\\\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~\mathsf{cotg^{n}x=\dfrac{1}{tg^{n}x}=\bigg(\dfrac{1}{tg\,x}\bigg)^{n}}$
________________________

Sabemos que

$\bullet\,\,\mathsf{tg^{2}x+1=sec^{2}x}\\\\\bullet\,\,\mathsf{cotg^{2}x+1=cossec^{2}x}$

Então:

$\mathsf{sec^{2}x+cossec^{2}x=(1+tg^{2}x)+(1+cotg^{2}x)}\\\\\\\mathsf{sec^{2}x+cossec^{2}x=2+tg^{2}x+cotg^{2}x}\\\\\\\mathsf{sec^{2}x+cossec^{2}x=2+tg^{2}x+\dfrac{1}{tg^{2}x}}\\\\\\\mathsf{sec^{2}x+cossec^{2}x=\dfrac{2tg^{2}x+tg^{4}x+1}{tg^{2}x}}\\\\\\\mathsf{sec^{2}x+cossec^{2}x=\dfrac{tg^{4}x+2tg^{2}x+1}{tg^{2}x}}\\\\\\\mathsf{sec^{2}x+cossec^{2}x=\dfrac{\big(tg^{2}x\big)^{2}+2\cdot(tg^{2}x)\cdot1+1^{2}}{tg^{2}x}}$

Note que temos a expansão de $\mathsf{\big(tg^{2}x+1\big)^{2}}$ no numerador, então:

$\mathsf{sec^{2}x+cossec^{2}x=\dfrac{(tg^{2}x+1)^{2}}{tg^{2}x}}$

Usando novamente $\mathsf{sec^{2}x=1+tg^{2}x}$:

$\mathsf{sec^{2}x+cossec^{2}x=\dfrac{\big(sec^{2}x\big)^{2}}{tg^{2}x}}\\\\\\\mathsf{sec^{2}x+cossec^{2}x=sec^{4}x\cdot cotg^{2}x}\\\\\\\mathsf{sec^{2}x+cossec^{2}x=\dfrac{1}{cos^{4}x}\cdot\dfrac{cos^{2}x}{sen^{2}x}}\\\\\\\mathsf{sec^{2}x+cossec^{2}x=\dfrac{1}{cos^{2}x}\cdot\dfrac{1}{sen^{2}x}}\\\\\\\boxed{\boxed{\mathsf{sec^{2}x+cossec^{2}x=sec^{2}x\cdot cossec^{2}x}}}$