Question

Answer asap ill mark brainliest

Answer 1

Please refer the attached image for fig.

Answer

$secA \: cosecC +cosecA \: secC \\ = 2 \times 2 + \frac{2}{ \sqrt{3} } \times \frac{2}{ \sqrt{3} } \\ = 4 + \frac{4}{3} \\ = \frac{12 + 4}{3} = \\ = \frac{16}{3}$

Steps and explanation

$\sin\theta = \frac{opp}{hyp} \\ cos\theta = \frac{adj}{hyp} \\ tan\theta = \frac{opp}{adj}$

$\theta = A(i.e. \: angleA)$

We have

$cotA = \frac{1}{ \sqrt{3} }$

As

$tanA = \frac{1}{cotA}$

So,

$tanA = \sqrt{3} \\ opp = \sqrt{3} \: and \: adj = 1$

BY Pythagoras theorem

${hyp}^{2} = {opp}^{2} + {adj}^{2} \\ {hyp}^{2} = { (\sqrt{3} )}^{2} + {1}^{2} \\ hyp = \sqrt{3 + 1} = \sqrt{4} = 2$

So

$sinA = \frac{ \sqrt{3} }{2} \\ inverse \: of \: sin \: is \: cosec \\ = >cosecA = \frac{2}{ \sqrt{3} } \\ cosA = \frac{1}{2} \\ similarly \\ cosA = \frac{1}{secA} \\ secA = 2$

Similarly we can find

$cosecC = 2 \\ and \\ secC = \frac{2}{ \sqrt{3} }$