Question

Simplify

Answer 1

\[\frac{ \sqrt{3} }{ 2 }+[\frac{ 1-\sqrt{3} }{ \sqrt{3} }*\frac{ \sqrt{3} }{ \sqrt{3+1} }]\]

Answer 2

ANSWER = \[\frac{ 3\sqrt{3}-4 }{ 2 }\]

Answer 3

isn't that sqrt 3+1 instead of sqrt(3+1) ??

Answer 4

I am postin the real question

Answer 5

\[\sin(\sin^-1\frac{ \sqrt{3} }{ 2 }+\cos^-1\frac{ 1 }{ 2 })+\tan(\sin^-1\frac{ 1 }{ 2 }+\tan^-1(-1))\]

Answer 6

i got till the step I've posted first

Answer 7

i am calculating using my calculator it is giving the right answer but when i do it myself i am doing it somewhere wrong

Answer 8

sin (60+60) = sin 120 = sqrt 3 /2 = first term = correct.
tan (30-45) = tan (-15) ... hmm

Answer 9

how did you go about finding tan 15 ?

Answer 10

i calculated in radians

Answer 11

show your work, i'll spot the error.

Answer 12

okay wait

Answer 13

i am stuck here how to add these

Answer 14

\[\frac{ \sqrt{3} }{ 2 }+\frac{ 1-\sqrt{3} }{ \sqrt{3}+1 }\]

Answer 15

for 2nd term, multiply and divide by sqrt 3-1

Answer 16

u mean conjugate??

Answer 17

yup.

Answer 18

don't we do it for denominator??

Answer 19

sqrt3 +1

Answer 20

in denominator of 2nd term, (sqrt 3 +1 ) (sqrt 3-1) =.... ?

Answer 21

in numerator of 2nd term, (1-sqrt 3) (sqrt 3-1) =.... ?

Answer 22

can u show me how to get it done

Answer 23

i am doing it wrong again i think

Answer 24

\(\large \frac{ \sqrt{3} }{ 2 }+\frac{ 1-\sqrt{3} }{ \sqrt{3}+1 } \times \frac{\sqrt3-1}{\sqrt3-1} = \)

Answer 25

i did it

Answer 26

:)

Answer 27

\[\frac{ \sqrt{3} }{ 2 } + \frac{ 2\sqrt{3}-4 }{ 2 }\]

Answer 28

\[\frac{ 3\sqrt{3}-4 }{ 2 }\]

Answer 29

yes....