Solving Proving Identities - Trig Chapter Problems

[paulmdrdo1]

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$

 

[Nono713]

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: $(a + b)^2 = a^2 + 2ab + b^2$.

 

[paulmdrdo1]

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?

 

[Nono713]

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.

 

[kaliprasad]

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

 

[paulmdrdo1]

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

 

[soroban]

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