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ahgua
2005-08-26 23:32:33

Prove that (sec^2 x + 2 tan x)/ (1 - tan^2 x) = (cos x + sin x)/(cos x - sin x)
Hence solve the equation (sin x +cos x)/(sin x - cos x) = tan x for 0 < x < 2pi.

ganesh
2005-08-26 22:09:31

cosecA + cotA = 3
To find cosecA - cotA,
one should remember,
cosec²A - cot²A = 1
From this, we get
(cosecA+ cotA)(cosecA - cotA)=1
Given cosecA + cotA = 3,
3(cosecA-cotA)=1
cosecA-cotA=1/3

cosecA + cotA + cosecA-cotA = 3+1/3 = 4/3
2cosecA = 4/3
cosecA=2/3
sinA=3/2 funny

ahgua
2005-08-26 20:41:34

thanks ajp3. But I got another question on trigo...

Given cosec A + cot A = 3

and thus evalute cosec A -cot A

and find cos A

ajp3
2005-08-26 11:31:05

ahgua, do not worry about dividing by cos x in the (correct) sol'n given by wcy; sin x/cos x is defined as tan x, so we do not need to worry about division by zero, the dfn of tan x already forbids it!

another way to show x=0 is a sol'n was already given by wcy in that soln; the sin x = 0 half of the soln...

ganesh
2005-08-24 22:16:35

You are right!
When x=0,
2sinx.cosx + cos^2x = 1
But how do we get a solution x=0?
2sinx.cosx + cos^2x = 1
sin2x + cos^2x = 1,
sin2x = 1-cos^2x
sin2x = sin^2x
this would be true only when x=0
But is there any (other)mathematical way it can be shown that x=0 is a solution?

ahgua
2005-08-24 20:59:24

#### wcy wrote:

sin x=2 cos x
tan x=2
x=...

Are you sure we can divide sin x by cos x?
what if cos x = 0 ...
then, the answer will be undefined!

wcy
2005-08-24 20:41:28

1=cos^2 x+sin^2 x
then, group all the sin terms...

2 sinx cosx +cos^2x=cos^2x +sin^2 x
2 sinx cosx=sin^2 x
sin^2 x-2 sinx cos x=0
sin x(sin x-2 cos x)=0
sin x=0 or sin x=2 cos x
x=0,etc..
or tan x=2
x=...

ahgua
2005-08-24 20:14:00

Solve this:

2sinx.cosx + cos^2x = 1

mathsyperson
2005-08-23 22:24:41

Making a right-angled triangle out of AB, the line going down from B and the line going right from A, we have one angle and one length. The 'opposite' side is 0.8m, the angle is 60° and we want to find the 'adjacent' side. tanθ=o/a, so tan 60=0.8/a, so a=0.8/tan 60=0.461... There are 21 of these, so 9.6995... of the 10m is being taken up by them. That leaves 0.3005... for CD. There are 2 of CD, so they will get 0.1503... each.

Making another right-angled triangle out of CD and the corner, we now know the side length and we want to know the hypotenuse. cosθ=a/h, so cos 60=0.1503.../h, so h=0.1503.../cos 60=0.30m to the nearest cm, which is length CD.

The outer frame has two lengths and two widths on it, so in total it is 10+10+0.8+0.8=21.6m long and one length AB is 0.8/sin 60=0.9238...m, so 21 of them are 19.40m to the nearest cm.

Titus
2005-08-23 21:44:53

The structure below measures 10m in length by 0.8m in width. In total, there are 21 diagonal pieces exactly like AB and two extra pieces- CD. Pieces CD are parallel to pieces AB. Pieces AB are inclined at 60 degrees from the horizontal.

After that i need to find the total length of the out frame and all 21 pieces of AB.

Any help is greatly appreciated.

Note: the vertical lines (apart from those on the outer frame) are not a part of the actual structure. They are merely usaed to form triangles so certain lengths can be calculated.

thanks.