Question

I have been asked to show that
1-sin 60 deg 1-tan 60 deg
----------- = -----------
cos 60 deg 1+tan 60 deg

Answer 1

But I am getting LHS = 2 - sqrt(3)

but

RHS = -2 + sqrt(3)

Answer 2

Where have I gone wrong??

Answer 3

maybe it's because you substituted the real values instead of proving...

Answer 4

@akitav yours becomes -2 + sqrt 3 too....

Answer 5

@HappySoul maybe they just aren't equal....

Answer 6

No I did not, I took -2 common in the numerator and cancelled it with the -2 in denominator

Answer 7

I think you left out a '-' in your original problem statement.

Answer 8

No, I have double checked the problem....

Answer 9

screen shot

Answer 10

you mean (1-sin60)(1-tan60)=cos60(1+tan60) ???

Answer 11

Not a very good pic but u can see the question.....

Answer 12

ok wait a minute

Answer 13

I feel there is some mistake in the question itself.....

Answer 14

I think it should be

\[(\frac{ 1-\sin 60^{0} }{ \cos60^{0}}) equals -(\frac{ 1-\tan60^{0} }{ 1+\tan60^{0} }) \]

Answer 15

damn.... i thought what is the problem..

Answer 16

is my thinking as above correct??

Answer 17

try by doing cross multiplication

Answer 18

it is not calculate in number.. its purpose is using trigonometrical function eqation.

Answer 19

try convert tan to sin and cos.

Answer 20

Since a specific angle is given, we have to use the value of the angle here (as per examples in the book).......
Pls try and solve by the method suggested by you, if you can get the answer, pls post it...thanks.....☺

Answer 21

ok i am doing it.

Answer 22

may be question is wronng pleasse check it twice.

Answer 23

yeah.. i am really stuck in proving

Answer 24

\[\frac{ 1-\sin60 }{ \cos60 }=\frac{ \cos60 - \sin60 }{ \cos60 + \sin60 } \]

Answer 25

what can i do later...