SOLUTION: prove this trig identity (5 marks):
tanx-1 = (sin^2x-cos^2x)/(sinxcosx+cos^2x)
I've wasted a good 8 pages trying to work this out haha. thanks in advance!
Algebra.Com
Question 199537: prove this trig identity (5 marks):
tanx-1 = (sin^2x-cos^2x)/(sinxcosx+cos^2x)
I've wasted a good 8 pages trying to work this out haha. thanks in advance!
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
-1=\frac{\sin^2(x)-\cos^2(x)}{\sin(x)\cos(x)+\cos^2(x)})
... Start with the given equation.
-1=\frac{\left(\sin(x)+\cos(x)\right) \left( \sin(x)-\cos(x)\right)}{ \sin(x)\cos(x)+\cos^2(x)})
... Factor the numerator using the difference of squares formula
-1=\frac{\left(\sin(x)+\cos(x)\right) \left( \sin(x)-\cos(x)\right)}{ \cos(x)\left(\sin(x)+\cos(x)\right)})
... Factor out the GCF from the denominator
-1=\frac{\sin(x)-\cos(x)}{\cos(x)})
... Cancel out the common terms.
-1=\frac{\sin(x)}{\cos(x)}-\frac{\cos(x)}{\cos(x)})
... Break up the fraction.
-1=\tan(x)-\frac{\cos(x)}{\cos(x)})
... Replace with
-1=\tan(x)-1)
... Reduce to get
So this verifies the identity.
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