MHB Solving Proving Identities - Trig Chapter Problems

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The discussion centers on solving two trigonometric identity problems that have proven challenging. The first problem involves simplifying the expression $\frac{1}{(\cos^{2}(x)-\sin^{2}(x))^2}-\frac{4\tan^{2}(x)}{(1-\tan^{2}(x))^2}=1$, with hints provided to expand and simplify using known identities. The second problem, $\frac{\tan(\theta)+\sec(\theta)-1}{\tan(\theta)-\sec(\theta)+1}=\frac{1+\sin(\theta)}{\cos(\theta)}$, suggests rewriting everything in terms of sine and cosine for easier manipulation. Participants share strategies and simplifications, ultimately leading to the conclusion that both problems can be resolved through established trigonometric identities. The thread emphasizes collaboration in tackling complex trigonometric proofs.
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Hey guys, I'm about to finish answering the chapter problems on my trigonometry books about proving Identities, there are 2 problems that made me scratch my head though. They are the last 2 problems left that I'm not yet able to verify. I would greatly appreciate it if you could lend me some hints to do these problems.

$\frac{1}{\left(\cos^{2}(x)-\sin^{2}(x)\right)^2}-\frac{4\tan^{2}(x)}{\left(1-\tan^{2}(x)\right)^2}=1$

$\frac{\tan(\theta)+\sec(\theta)-1}{\tan(\theta)-1\sec(\theta)+1}=\frac{1+\sin(\theta)}{\cos(\theta)}$

from here I have no idea what to do next. please help. Thanks!My attempt for prob1

$\frac{1}{\left(\cos^{2}(x)-\sin^{2}(x)\right)^2}-\frac{(4)\frac{\sin^{2}(x)}{\cos^{2}(x)}}{\left(1-\frac{\sin^{2}(x)}{\cos^{2}(x)}\right)^2}=1$$\frac{1-4\cos^{2}(x)\sin^{2}(x)}{\left(\cos^{2}(x)-\sin^{2}(x)\right)^2}=1$
 
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paulmdrdo said:
Hey guys, I'm about to finish answering the chapter problems on my trigonometry books about proving Identities, there are 2 problems that made me scratch my head though. They are the last 2 problems left that I'm not yet able to verify. I would greatly appreciate it if you could lend me some hints to do these problems.

$\frac{1}{\left(\cos^{2}(x)-\sin^{2}(x)\right)^2}-\frac{4\tan^{2}(x)}{\left(1-\tan^{2}(x)\right)^2}=1$

$\frac{\tan(\theta)+\sec(\theta)-1}{\tan(\theta)-1\sec(\theta)+1}=\frac{1+\sin(\theta)}{\cos(\theta)}$

from here I have no idea what to do next. please help. Thanks!My attempt for prob1

$\frac{1}{\left(\cos^{2}(x)-\sin^{2}(x)\right)^2}-\frac{(4)\frac{\sin^{2}(x)}{\cos^{2}(x)}}{\left(1-\frac{\sin^{2}(x)}{\cos^{2}(x)}\right)^2}=1$$\frac{1-4\cos^{2}(x)\sin^{2}(x)}{\left(\cos^{2}(x)-\sin^{2}(x)\right)^2}=1$

Hint: you get:

$$\frac{1-4\cos^{2}(x)\sin^{2}(x)}{\left(\cos^{2}(x)-\sin^{2}(x)\right)^2}=1$$

Hence:

$$1-4\cos^{2}(x)\sin^{2}(x)=\left(\cos^{2}(x)-\sin^{2}(x)\right)^2$$

Now expand the RHS to get:

$$1-4\cos^{2}(x)\sin^{2}(x)= \cos^4(x) - 2 \cos^2(x) \sin^2(x) + \sin^4(x)$$

And simplify:

$$1-2\cos^{2}(x)\sin^{2}(x)= \cos^4(x) + \sin^4(x)$$

$$\cos^4(x) + 2\cos^{2}(x)\sin^{2}(x) + \sin^4(x) = 1$$

Does that look like anything you know? Hint: $(a + b)^2 = a^2 + 2ab + b^2$.
 
Bacterius said:
Hint: you get:

$$\frac{1-4\cos^{2}(x)\sin^{2}(x)}{\left(\cos^{2}(x)-\sin^{2}(x)\right)^2}=1$$

Hence:

$$1-4\cos^{2}(x)\sin^{2}(x)=\left(\cos^{2}(x)-\sin^{2}(x)\right)^2$$

Now expand the RHS to get:

$$1-4\cos^{2}(x)\sin^{2}(x)= \cos^4(x) - 2 \cos^2(x) \sin^2(x) + \sin^4(x)$$

And simplify:

$$1-2\cos^{2}(x)\sin^{2}(x)= \cos^4(x) + \sin^4(x)$$

$$\cos^4(x) + 2\cos^{2}(x)\sin^{2}(x) + \sin^4(x) = 1$$

Does that look like anything you know? Hint: $(a + b)^2 = a^2 + 2ab + b^2$.

Yes! $\left(\cos^{2}(x)+\sin^{2}(x)\right)^2=1$

$1=1$

How about the second problem?
 
paulmdrdo said:
How about the second problem?

Try multiplying out the fractions, and rewriting everything in terms of $\sin$ and $\cos$. To get you started:

$$\frac{\tan(\theta)+\sec(\theta)-1}{\tan(\theta)-\sec(\theta)+1}=\frac{1+\sin(\theta)}{\cos(\theta)}$$

$$(\cos(\theta)) (\tan(\theta)+\sec(\theta)-1)=(\tan(\theta)-\sec(\theta)+1) (1+\sin(\theta))$$

$$(\cos(\theta)) (\frac{\sin(\theta)}{\cos(\theta)}+\frac{1}{\cos(\theta)}-1)=(\frac{\sin(\theta)}{\cos(\theta)}- \frac{1}{\cos(\theta)}+1) (1+\sin(\theta))$$

Notice that a bunch of stuff simplifies right off the bat in the LHS. Now try expanding the RHS, and see where that gets you! Good luck.
 
Bacterius said:
Try multiplying out the fractions, and rewriting everything in terms of $\sin$ and $\cos$. To get you started:

$$\frac{\tan(\theta)+\sec(\theta)-1}{\tan(\theta)-\sec(\theta)+1}=\frac{1+\sin(\theta)}{\cos(\theta)}$$

$$(\cos(\theta)) (\tan(\theta)+\sec(\theta)-1)=(\tan(\theta)-\sec(\theta)+1) (1+\sin(\theta))$$

$$(\cos(\theta)) (\frac{\sin(\theta)}{\cos(\theta)}+\frac{1}{\cos(\theta)}-1)=(\frac{\sin(\theta)}{\cos(\theta)}- \frac{1}{\cos(\theta)}+1) (1+\sin(\theta))$$

Notice that a bunch of stuff simplifies right off the bat in the LHS. Now try expanding the RHS, and see where that gets you! Good luck.

I love this problem as

$\dfrac{\tan(\theta)+\sec(\theta)-1}{\tan(\theta)-\sec(\theta)+1}$
= $\dfrac{\tan(\theta)+\sec(\theta)-(\sec^2(\theta)- \tan ^2 (\theta)}{\tan(\theta)-\sec(\theta)+1}$
= $\dfrac{(\tan(\theta)+\sec(\theta))( 1 -(\sec(\theta)- \tan (\theta))}{\tan(\theta)-\sec(\theta)+1}$
= $\dfrac{(\tan(\theta)+\sec(\theta))( 1 -\sec(\theta)+ \tan (\theta))}{\tan(\theta)-\sec(\theta)+1}$
= $\tan(\theta)+\sec(\theta)$
= $\dfrac{\sin(\theta)}{\cos(\theta)}+\dfrac{1}{\cos(\theta)}$
= $\dfrac{1+\sin(\theta)}{\cos(\theta)}$

I love it because replacing 1 by $\sec^2(\theta)- \tan ^2 (\theta)$ simplifies a lot
 
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Bacterius said:
Try multiplying out the fractions, and rewriting everything in terms of $\sin$ and $\cos$. To get you started:

$$\frac{\tan(\theta)+\sec(\theta)-1}{\tan(\theta)-\sec(\theta)+1}=\frac{1+\sin(\theta)}{\cos(\theta)}$$

$$(\cos(\theta)) (\tan(\theta)+\sec(\theta)-1)=(\tan(\theta)-\sec(\theta)+1) (1+\sin(\theta))$$

$$(\cos(\theta)) (\frac{\sin(\theta)}{\cos(\theta)}+\frac{1}{\cos(\theta)}-1)=(\frac{\sin(\theta)}{\cos(\theta)}- \frac{1}{\cos(\theta)}+1) (1+\sin(\theta))$$

Notice that a bunch of stuff simplifies right off the bat in the LHS. Now try expanding the RHS, and see where that gets you! Good luck.
Again I end up getting $\sin^{2}(\theta)+\cos^{2}(\theta)=1$
 
Hello, paulmdrdo!

$\frac{1}{\left(\cos^2x -\sin^2x\right)^2}-\frac{4\tan^2x}{\left(1-\tan^2x\right)^2}\;=\;1$
We are expected to know these three identities:

. . \cos2\theta \:=\:\cos^2\!\theta - \sin^2\!\theta

. . \tan2\theta \:=\:\frac{2\tan\theta}{1-\tan^2\!\theta}

$\quad \sec^2\!\theta - \tan^2\!\theta \:=\:1$The first fraction is: $\:\dfrac{1}{(\cos^2\!x - \sin^2\!x)^2} \:=\:\dfrac{1}{\cos^2\!2x} \:=\:\sec^2\!2x$

The second fraction is: $\:\left(\dfrac{2\tan x}{1-\tan^2\!x}\right)^2 \:=\:\tan^2\!2x$

And we have: $\:\dfrac{1}{\left(\cos^2\!x -\sin^2\!x\right)^2}-\dfrac{4\tan^2\!x}{\left(1-\tan^2\!x\right)^2} \;=\; \sec^2\!x - \tan^2\!x \; =\;1$
 

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