Hence solve the equation (sin x +cos x)/(sin x - cos x) = tan x for 0 < x < 2pi.]]>

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

Given cosec A + cot A = 3

and thus evalute cosec A -cot A

and find cos A

]]>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...

]]>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?]]>

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!

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=...

2sinx.cosx + cos^2x = 1

]]>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.

]]>Please help mefind the length (in millimetres) of piece CD as i already know how to find the length of AB.

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.

]]>