Question

hey! plz help
$\frac{cos \: 45}{sec \: 30 \: + \: cosec \: 30 }$​

Answer 1

cos45/sec30+cosec30

Explanation:

cos 45 + 30 + cosec 30 is equal to 1 by root 2 because cos 45 1 by root 2 / 2 by root 3 + 2 = 1 by root 2 / 2 + 2 root 3 divided by root 3 next step 1 by root 2 into root 3 by 2 ( 1 + root 3 ) root3/2root2(1+root3)

sorry root cannot expressed numerically

Answer 2

Here's your answer :-

$= > \frac{cos \: 45}{sec \: 30 + cosec \: 30}$

$= > \frac{ \frac{1}{ \sqrt{2}}}{ \frac{2}{ \sqrt{3}}+2} = \frac{ \frac{1}{ \sqrt{2} } }{\frac{2 + 2 \sqrt{2} }{ \sqrt{3} } }$

• Rationalize by : 1/√2

$= > \frac{1}{ \sqrt{2}} \times \frac{ \sqrt{2} }{ \sqrt{2}} = \frac{ \sqrt{2} }{2}$

$= > \frac{ \sqrt{2} }{2} \div \frac{ \sqrt{3} }{2 + 2 \sqrt{3} }$

• Now, rationalize by : 1-√3

$= > \frac{ \sqrt{2} }{2} \times \frac{ \sqrt{3} }{2} \times ( \frac{1}{1 - \sqrt{3}})$

$= > \frac{ \sqrt{6} }{4} \times \frac{(1 - \sqrt{3})}{1 - 3}$

$= > \frac{ \sqrt{6}(1 - \sqrt{3})}{ - 8}$

$= > \frac{ \sqrt{6} - \sqrt{18} }{ - 8} = \frac{ - ( \sqrt{ 18} + \sqrt{6})}{ - (8)}$

$= > \frac{3 \sqrt{2} - \sqrt{6} }{8}$

Hope it helps!

~ A.R.M.Y ~