Question

Find the exact value of the expression.

tan(cos^-1(root3/2))

Answer 1

@mathslover

Answer 2

is this your question? ? :
\[\huge{tan(cos^{-1}(\sqrt{\frac{3}{2}}))}\]

Answer 3

was that your question?

Answer 4

yeah

Answer 5

cos^-1 represents inverse or just 1/cos ?

Answer 6

inverse

Answer 7

so let :
\[\huge{tan(cos^{-1}(\sqrt{\frac{3}{2}}))=tan(x) }\]
where x = cos^{-1}(root(3/2))

Answer 8

ok till now?

Answer 9

yeah

Answer 10

so as we know that

cos x = -\(\large{\frac{\sqrt{3}}{2}}\)

cos^2x = ? can u tell me

Answer 11

what is the need of cos^2x? @mathslover

Answer 12

there is the need you just tell me what will be cos^2 x if cos x = - root3 /2

Answer 13

u there @KingAditya ?

Answer 14

yeah

Answer 15

answer my question aditya...

what will be cos^2 x if cos x = - root3 /2

Answer 16

wait

Answer 17

ok

Answer 18

@KingAditya r u confused?
tell me :
\[\large{\textbf{what is :}(\frac{-\sqrt{3}}{2})^2}\]

Answer 19

i think it might be 3/4

Answer 20

cos^2=3/4

Answer 21

cos^2x=3/4

Answer 22

oh great ..


now cos^2 x = 1 - sin^2 x

so : 3/4 = 1 - sin^2 x

1 - 3/4 = sin^2 x
1/4 = sin^2 x

sin x = 1/2 right?

Answer 23

\[\large{tan x = \frac{sin x}{cos x}}\]
\[\large{tan x = \frac{-1}{\sqrt{3}}}\]

Answer 24

that is ur answer.. got it?

Answer 25

1/sqrt(3)??

Answer 26

-1/sqrt{3} sorry for that mistake

Answer 27

-??

Answer 28

yes -1/sqrt{3}

Answer 29

it might be +ve if theta is in 3rd quadrant

Answer 30

u mean in T quadrant??

Answer 31

tan is +ve in third quadrant

Answer 32

got it?

Answer 33

@mathslover , he said cos^-1 represents the inverse, so it is arccos root3/2 which is Pi/6 . And tan(Pi/6) is 1/root3 = root3/3. That's it.
@KingAditya

Answer 34

oh yes right myko