Question

Sabendo que tg x= 4/3, calcule 1/cos2x cossec2 x

Answer 1

Cossec²x=1+cotg²x ---é igual a o inverso da tgx
cossec²x =1+( $\frac{3}{4}$ )²       ----> cossec²x =$\frac{25}{16}$
cossecx=$\frac{5}{4}$

sen²x+cos²x=1
$( \frac{4}{5} )²$ + cos²x =1 ----->cos²x =$\frac{9}{25}$ --->cosx =$\frac{3}{5}$

Answer 2

$\dfrac{1}{cos~2x\cdot cossec~2x}=\dfrac{1}{cos~2x\cdot\left(\frac{1}{sen~2x}\right)}\\\\\\\dfrac{1}{cos~2x\cdot cossec~2x}=\dfrac{sen~2x}{cos~2x}\\\\\\\dfrac{1}{cos~2x\cdot cossec~2x}=tg~2x$

Achando uma expressão pra tg 2x:

$tg~2x=tg~(x+x)\\\\\\tg~2x=\dfrac{tg~x+tg~x}{1-tg~x\cdot tg~x}=\dfrac{2tg~x}{1-tg^{2}x}$

Substituindo:

$\dfrac{1}{cos~2x\cdot cossec~2x}=\dfrac{2tg~x}{1-tg^{2}x}\\\\\\\dfrac{1}{cos~2x\cdot cossec~2x}=\dfrac{2\left(\frac{4}{3}\right)}{1-\left(\frac{4}{3}\right)^{2}}\\\\\\\dfrac{1}{cos~2x\cdot cossec~2x}=\dfrac{\left(\frac{8}{3}\right)}{1-\left(\frac{16}{9}\right)}\\\\\\\dfrac{1}{cos~2x\cdot cossec~2x}=\dfrac{\left(\frac{8}{3}\right)}{\left(\frac{9-16}{9}\right)}\\\\\\\dfrac{1}{cos~2x\cdot cossec~2x}=\dfrac{\left(\frac{8}{3}\right)}{\left(\frac{-7}{9}\right)}$

Fazendo a divisão:

$\dfrac{1}{cos~2x\cdot cossec~2x}=\dfrac{8}{3}\cdot\dfrac{-9}{7}\\\\\\\dfrac{1}{cos~2x\cdot cossec~2x}=\dfrac{8}{1}\cdot\dfrac{-3}{7}\\\\\\\boxed{\boxed{\dfrac{1}{cos~2x\cdot cossec~2x}=-\dfrac{24}{7}}}$