Verify the idenity: (1-tanx)(1-cotx)=2-sec x csc x
H haze New member Joined Jul 23, 2008 Messages 1 Jul 23, 2008 #1 Verify the idenity: (1-tanx)(1-cotx)=2-sec x csc x
pka Elite Member Joined Jan 29, 2005 Messages 11,995 Jul 23, 2008 #2 Change both tangent and cotangent to functions of sine and cosine. Do it and show us what you get.
D Deleted member 4993 Guest Jul 26, 2008 #3 haze said: Verify the idenity: (1-tanx)(1-cotx)=2-sec x csc x Click to expand... Since few days gone by: [1 − tan(x)]⋅[1 −cot(x)]\displaystyle [1\, - \, \tan(x)]\cdot[1 \, - \cot(x)][1−tan(x)]⋅[1−cot(x)] using foil 1 + [tan(x)]⋅[cot(x)] −[tan(x) + cot(x)]\displaystyle 1\, + \, [\tan(x)]\cdot[cot(x)] \, - [\tan(x) \, + \, \cot(x)]1+[tan(x)]⋅[cot(x)]−[tan(x)+cot(x)] 1 + 1 −[sin(x)cos(x) + cos(x)sin(x)]\displaystyle 1\, + \, 1 \, - [\frac{\sin(x)}{cos(x)} \, + \, \frac{\cos(x)}{\sin(x)}]1+1−[cos(x)sin(x)+sin(x)cos(x)] 2 −[sin2(x) + cos2(x)cos(x)⋅sin(x)]\displaystyle 2 \, - [\frac{\sin^2(x) \, + \, \cos^2(x)}{\cos(x) \cdot \sin(x)}]2−[cos(x)⋅sin(x)sin2(x)+cos2(x)] 2 −[1cos(x) ⋅ 1sin(x)]\displaystyle 2 \, - [\frac{1}{cos(x)} \, \cdot \, \frac{1}{\sin(x)}]2−[cos(x)1⋅sin(x)1] and so on...
haze said: Verify the idenity: (1-tanx)(1-cotx)=2-sec x csc x Click to expand... Since few days gone by: [1 − tan(x)]⋅[1 −cot(x)]\displaystyle [1\, - \, \tan(x)]\cdot[1 \, - \cot(x)][1−tan(x)]⋅[1−cot(x)] using foil 1 + [tan(x)]⋅[cot(x)] −[tan(x) + cot(x)]\displaystyle 1\, + \, [\tan(x)]\cdot[cot(x)] \, - [\tan(x) \, + \, \cot(x)]1+[tan(x)]⋅[cot(x)]−[tan(x)+cot(x)] 1 + 1 −[sin(x)cos(x) + cos(x)sin(x)]\displaystyle 1\, + \, 1 \, - [\frac{\sin(x)}{cos(x)} \, + \, \frac{\cos(x)}{\sin(x)}]1+1−[cos(x)sin(x)+sin(x)cos(x)] 2 −[sin2(x) + cos2(x)cos(x)⋅sin(x)]\displaystyle 2 \, - [\frac{\sin^2(x) \, + \, \cos^2(x)}{\cos(x) \cdot \sin(x)}]2−[cos(x)⋅sin(x)sin2(x)+cos2(x)] 2 −[1cos(x) ⋅ 1sin(x)]\displaystyle 2 \, - [\frac{1}{cos(x)} \, \cdot \, \frac{1}{\sin(x)}]2−[cos(x)1⋅sin(x)1] and so on...
