# Trigonometry

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Figure 1 shows the right-angled triangles ABC, ABD and BDC, with AB=1 and ÐBAD=q .

Prove that 1+tan^2q =sec^2q .

Prove that 1+tan^2q =sec^2q .

Last edited by Ljm2001; 1 year ago

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#2

(Original post by

Figure 1 shows the right-angled triangles ABC, ABD and BDC, with AB=1 and ÐBAD=q .

Prove that 1+tan^2q =sec^2q .

**Ljm2001**)Figure 1 shows the right-angled triangles ABC, ABD and BDC, with AB=1 and ÐBAD=q .

Prove that 1+tan^2q =sec^2q .

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I understand how to prove it without using the triangle, but I clearly need to use it in some way, that is what is confusing me

(Original post by

Any thoughts/idea/progress?

**ghostwalker**)Any thoughts/idea/progress?

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#4

(Original post by

I understand how to prove it without using the triangle, but I clearly need to use it in some way, that is what is confusing me

**Ljm2001**)I understand how to prove it without using the triangle, but I clearly need to use it in some way, that is what is confusing me

Looking at the triangle(s), we could use Pythagoras either on ABD or on ABC as our starting point, though it's only one step removed from the trig identity.

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thanks! that would work, but then why would they show both triangles?

(Original post by

I guess you're refering to the identity.

Looking at the triangle(s), we could use Pythagoras either on ABD or on ABC as our starting point, though it's only one step removed from the trig identity.

**ghostwalker**)I guess you're refering to the identity.

Looking at the triangle(s), we could use Pythagoras either on ABD or on ABC as our starting point, though it's only one step removed from the trig identity.

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#6

(Original post by

thanks! that would work, but then why would they show both triangles?

**Ljm2001**)thanks! that would work, but then why would they show both triangles?

The only thing I can think of is they want you to avoid using Pythagoras, and work with the similar triangles - in which case they should have specified that. The exact way to do that eludes me at present though - assuming it is feasible.

Last edited by ghostwalker; 1 year ago

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#7

(Original post by

Actually there are three similar triangles there. Everything relating to D is redundant if we use Pythagoras.

The only thing I can think of is they want you to avoid using Pythagoras, and work with the similar triangles - in which case they should have specified that. The exact way to do that eludes me at present though - assuming it is feasible.

**ghostwalker**)Actually there are three similar triangles there. Everything relating to D is redundant if we use Pythagoras.

The only thing I can think of is they want you to avoid using Pythagoras, and work with the similar triangles - in which case they should have specified that. The exact way to do that eludes me at present though - assuming it is feasible.

I was looking for solutions online and couldn't find any so I decided to actually try doing it myself and somehow got the answer

https://photos.google.com/share/AF1Q...VhckVZWEpMb1ZR

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#8

I acc have a mark scheme answer for this! But it's written down and it's at school

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#9

(Original post by

I acc have a mark scheme answer for this! But it's written down and it's at school

**Afrrcyn**)I acc have a mark scheme answer for this! But it's written down and it's at school

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#11

(Original post by

im 1 year late but I figured out how to do this question lol

I was looking for solutions online and couldn't find any so I decided to actually try doing it myself and somehow got the answer

https://photos.google.com/share/AF1Q...VhckVZWEpMb1ZR

**jacob1140**)im 1 year late but I figured out how to do this question lol

I was looking for solutions online and couldn't find any so I decided to actually try doing it myself and somehow got the answer

https://photos.google.com/share/AF1Q...VhckVZWEpMb1ZR

It should just be a couple of lines, one to write down Pythagoras on ABC, the other to replace the lengths with the equivalent trig terms. As noted above, I can't see how the triangle split makes it any simpler (can't get much simpler than a couple of lines) unless they want you to imagine ADB as the starting point, then consider adding BDC to it to get ABC. You could then note dividing by cos on ABD does the "same" as adding BDC to get ABC. You're constructing a new, similar triangle with the desired length property.

You could use similar triangles without assuming Pythagoras (noted above and basically proving Pythagoras) as

AC = AD + DC

and get AD and DC using similar triangles to ABC. A bit longer though and more complex.

Last edited by mqb2766; 1 month ago

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#12

(Original post by

Is the first line correct? You seem to have leg > hypotenuse? Also x isn't defined?

It should just be a couple of lines, one to write down Pythagoras on ABC, the other to replace the lengths with the equivalent trig terms. As noted above, I can't see how the triangle split makes it any simpler (can't get much simpler than a couple of lines) unless they want you to imagine ADB as the starting point, then consider adding BDC to it to get ABC. You could then note dividing by cos on ABD does the "same" as adding BDC to get ABC. You're constructing a new, similar triangle with the desired length property.

You could use similar triangles without assuming Pythagoras (noted above and basically proving Pythagoras) as

AC = AD + DC

and get AD and DC using similar triangles to ABC. A bit longer though and more complex.

**mqb2766**)Is the first line correct? You seem to have leg > hypotenuse? Also x isn't defined?

It should just be a couple of lines, one to write down Pythagoras on ABC, the other to replace the lengths with the equivalent trig terms. As noted above, I can't see how the triangle split makes it any simpler (can't get much simpler than a couple of lines) unless they want you to imagine ADB as the starting point, then consider adding BDC to it to get ABC. You could then note dividing by cos on ABD does the "same" as adding BDC to get ABC. You're constructing a new, similar triangle with the desired length property.

You could use similar triangles without assuming Pythagoras (noted above and basically proving Pythagoras) as

AC = AD + DC

and get AD and DC using similar triangles to ABC. A bit longer though and more complex.

However I did it again and found a solution lol wait a sec lemme post it

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#13

**mqb2766**)

Is the first line correct? You seem to have leg > hypotenuse? Also x isn't defined?

It should just be a couple of lines, one to write down Pythagoras on ABC, the other to replace the lengths with the equivalent trig terms. As noted above, I can't see how the triangle split makes it any simpler (can't get much simpler than a couple of lines) unless they want you to imagine ADB as the starting point, then consider adding BDC to it to get ABC. You could then note dividing by cos on ABD does the "same" as adding BDC to get ABC. You're constructing a new, similar triangle with the desired length property.

You could use similar triangles without assuming Pythagoras (noted above and basically proving Pythagoras) as

AC = AD + DC

and get AD and DC using similar triangles to ABC. A bit longer though and more complex.

heres the answer I got 100% overall marks but also suspected of cheating lol

Last edited by jacob1140; 1 month ago

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#14

(Original post by

https://photos.app.goo.gl/VUZLP3nMj1siHyWZA

heres the answer I got 100% overall marks but also suspected of cheating lol

**jacob1140**)https://photos.app.goo.gl/VUZLP3nMj1siHyWZA

heres the answer I got 100% overall marks but also suspected of cheating lol

Hard to comment too much on the cheating bit apart from to say your first post was "looking for solutions online" and you don't explain the working ... Maybe they have a point?

Last edited by mqb2766; 1 month ago

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#15

(Original post by

Looks ok, there are other solutions, some simpler. Don't know if you'd get full marks in an exam. There is little/no explanation of what you're doing or why. Even if it's just so the examiner marks it correctly or so you can check your own work, explanations help.

Hard to comment too much on the cheating bit apart from to say your first post was "looking for solutions online" and you don't explain the working ... Maybe they have a point?

**mqb2766**)Looks ok, there are other solutions, some simpler. Don't know if you'd get full marks in an exam. There is little/no explanation of what you're doing or why. Even if it's just so the examiner marks it correctly or so you can check your own work, explanations help.

Hard to comment too much on the cheating bit apart from to say your first post was "looking for solutions online" and you don't explain the working ... Maybe they have a point?

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#16

**mqb2766**)

Looks ok, there are other solutions, some simpler. Don't know if you'd get full marks in an exam. There is little/no explanation of what you're doing or why. Even if it's just so the examiner marks it correctly or so you can check your own work, explanations help.

Hard to comment too much on the cheating bit apart from to say your first post was "looking for solutions online" and you don't explain the working ... Maybe they have a point?

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#17

where did you get tan^2 from? and then you got a function with cos at the bottom?this question makes literall zero sense to me

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