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#### Pranav

##### Well-known member

- Nov 4, 2013

- 428

$$\frac{1}{\cos^210^{\circ}}+\frac{1}{\sin^220^{ \circ }}+\frac{1}{\sin^240^{\circ}}-\frac{1}{\cos^245^{\circ}}$$

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- #1

- Nov 4, 2013

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$$\frac{1}{\cos^210^{\circ}}+\frac{1}{\sin^220^{ \circ }}+\frac{1}{\sin^240^{\circ}}-\frac{1}{\cos^245^{\circ}}$$

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- Feb 7, 2012

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- Nov 4, 2013

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Thanks Opalg for your participation, your answer is correct.

Btw, is their any elementary solution to this? I wonder if the problem really involves the use of such complicated approach as it is a problem from one of my past test papers.

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- Feb 14, 2012

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- Nov 4, 2013

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Excellent!

But can you please explain how do you get the following:

$$1+\tan^210^{\circ}+\tan^270^{ \circ }+\tan^250^{\circ}=1+9\tan^230^{\circ}+6$$

Thanks!

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- #6

- Feb 14, 2012

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Sure!Excellent!

But can you please explain how do you get the following:

$$1+\tan^210^{\circ}+\tan^270^{ \circ }+\tan^250^{\circ}=1+9\tan^230^{\circ}+6$$

Thanks!

Do you know there is such an identity that tells us $\tan^2x^{\circ}+\tan^2(60-x)^{ \circ }+\tan^2(60+x)^{\circ}=9\tan^2(3x)^{\circ}+6$ ?

That is the exact identity that I used to translate the messy sum of three square terms of tangent functions into number!

But seriously though, I don't know for sure if we could just use that identity right away or we needed to prove it first before applying.

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- Nov 4, 2013

- 428

Nope, never heard of it, are there any more of identities of this kind? That looks very useful.Do you know there is such an identity that tells us $\tan^2x^{\circ}+\tan^2(60-x)^{ \circ }+\tan^2(60+x)^{\circ}=9\tan^2(3x)^{\circ}+6$ ?

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- Feb 14, 2012

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You're welcome,Thanks anemone!

Nope, never heard of it, are there any more of identities of this kind? That looks very useful.

Last edited:

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- #9

- Nov 4, 2013

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I am not sure what you meant there.yes, there are still a few more in my notebook, but, what would I get in return, hehehe...perhaps a cup of coffee?

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- Feb 14, 2012

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I was actually trying to make some fun here. I also meant I would like very much to share with you the other useful trigonometric identities that I know of but could I do that by tomorrow? Now it is pretty late here and I can barely keep my eyes open...I am not sure what you meant there.

Night night MHB!

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- #11

- Nov 4, 2013

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Please take your time and good night.I was actually trying to make some fun here. I also meant I would like very much to share with you the other useful trigonometric identities that I know of but could I do that by tomorrow? Now it is pretty late here and I can barely keep my eyes open...

Night night MHB!

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- #12

- Feb 14, 2012

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$\sin x^{\circ} \cdot \sin (60-x)^{\circ} \cdot \sin (60+x)^{\circ}=\dfrac{\sin 3x^{\circ}}{4}$

$\cos x^{\circ} \cdot \cos (60-x)^{\circ} \cdot \cos (60+x)^{\circ}=\dfrac{\cos 3x^{\circ}}{4}$

hence

$\tan x^{\circ} \cdot \tan (60-x)^{\circ} \cdot \tan (60+x)^{\circ}=\tan 3x^{\circ}$

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- #13

- Nov 4, 2013

- 428

Thanks a lot anemone!Pranav, here are the other trigonometric identities that I want to share with you:

$\sin x^{\circ} \cdot \sin (60-x)^{\circ} \cdot \sin (60+x)^{\circ}=\dfrac{\sin 3x^{\circ}}{4}$

$\cos x^{\circ} \cdot \cos (60-x)^{\circ} \cdot \cos (60+x)^{\circ}=\dfrac{\cos 3x^{\circ}}{4}$

hence

$\tan x^{\circ} \cdot \tan (60-x)^{\circ} \cdot \tan (60+x)^{\circ}=\tan 3x^{\circ}$