SOLUTION: Prove that Sin(theta)times Cos(theta) divided by Cos^2(theta)-Sin^2(theta)=Tan(theta) divided by 1-Tan(theta)
Show step by step how to change the left to equal the right side of e
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Question 109301This question is from textbook Algebra and Trigonomometry
: Prove that Sin(theta)times Cos(theta) divided by Cos^2(theta)-Sin^2(theta)=Tan(theta) divided by 1-Tan(theta)
Show step by step how to change the left to equal the right side of equasion
This question is from textbook Algebra and Trigonomometry
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
I'm going to use the double angle identities =2\sin(\theta)\cos(\theta))
, =\cos^{2}(\theta)-\sin^{2}(\theta))
, and =\frac{2\tan(\theta)}{1-\tan(\theta)})
Notice how we have \sin(\theta))
in the problem. So we must divide both sides of the identity by 2 to get =\sin(\theta)\cos(\theta))
\sin(\theta)}{\cos^{2}(\theta)-\sin^{2}(\theta)}&=\frac{\tan(\theta)}{1-\tan(\theta)} \\ \frac{\frac{1}{2}\sin(2\theta)}{\cos^{2}(\theta)-\sin^{2}(\theta)}&= \mbox{ Use the first identity listed}\\ \frac{\frac{1}{2}\sin(2\theta)}{\cos(2\theta)}&=\mbox{ Use the second identity listed}\\ \frac{1}{2}\tan(2\theta)&= \mbox{ Replace sine over cosine with tangent}\\ \frac{1}{2}\frac{2\tan(\theta)}{1-\tan(\theta)}&= \mbox{ Use the third identity listed}\\ \frac{\tan(\theta)}{1-\tan(\theta)}&= \mbox{ Simplify}\\ \end{align*})
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