Question

(UCPEL) A expressão 1-sec²x/1-csc²x é idêntica a:

a)cosx
b)sen²x
c)tan⁴x
d)tan²x

Answer 1

Olá Sara!

Para solucionar esse exercício, precisamos saber o seguinte:

$\\ \displaytyle \mathsf{(i) \qquad sin^2 x + \cos^2 x = 1 \Rightarrow \begin{cases} \mathsf{\sin^2 x = 1 - \cos^2 x \qquad (\ast)} \\ \mathsf{\cos^2 x = 1 - \sin^2 x \qquad (\ast\ast)} \end{cases}} \\\\ \mathsf{(ii) \qquad \sec x = \frac{1}{\cos x}} \\\\\\ \mathsf{(iii) \qquad cossec \ x = \frac{1}{\sin x}} \\\\\\ \mathsf{(iv) \qquad \tan x = \frac{\sin x}{\cos x}}$

Isto posto, fazemos:

$\\ \displaystyle \mathsf{\frac{1 - \sec^2 x}{1 - cossec^2 \ x} = \frac{1 - \frac{1}{\cos^2 x}}{1 - \frac{1}{\sin^2 x}} =} \\\\\\ \mathsf{= \left ( 1 - \frac{1}{\cos^2 x} \right ) \div \left ( 1 - \frac{1}{\sin^2 x} \right )} \\\\\\ \mathsf{= \left ( \frac{\cos^2 x - 1}{\cos^2 x} \right ) \div \left ( \frac{\sin^2 x - 1}{\sin^2 x} \right )} \\\\\\ \mathsf{= \left [ \frac{- \left (- \cos^2 x + 1 \right )}{\cos^2 x} \right ] \div \left [ \frac{- \left (- \sin^2 x + 1 \right )}{\sin^2 x} \right ]}$

$\\ \displaystyle \mathsf{= \left [ \frac{- \left (1 - \cos^2 x \right )}{\cos^2 x} \right ] \div \left [ \frac{- \left (1 - \sin^2 x \right )}{\sin^2 x} \right ]} \\\\\\ \mathsf{= \left [ \frac{- \left (sin^2 x \right )}{\cos^2 x} \right ] \div \left [ \frac{- \left (\cos^2 x \right )}{\sin^2 x} \right ]} \\\\\\ \mathsf{= \left [ \frac{- \left (sin^2 x \right )}{\cos^2 x} \right ] \cdot \left [ \frac{\sin^2 x}{- \left (\cos^2 x \right )} \right ]} \\\\\\ \mathsf{= \frac{- \sin^4 x}{- \cos^4 x}}$

$\\ \displaystyle \mathsf{= \frac{\sin^4 x}{\cos^4 x}} \\\\\\ \mathsf{= \left ( \frac{\sin x}{\cos x} \right )^4} \\\\\\ \boxed{\boxed{\mathsf{\frac{1 - \sec^2 x}{1 - cossec^2 \ x} = \tan^4 x}}}$