# Trigonometry

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#1
Figure 1 shows the right-angled triangles ABC, ABD and BDC, with AB=1 and ÐBAD=q .
Prove that 1+tan^2q =sec^2q .
Last edited by Ljm2001; 1 year ago
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1 year ago
#2
(Original post by 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 .
Any thoughts/idea/progress?
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#3
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 ghostwalker)
Any thoughts/idea/progress?
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1 year ago
#4
(Original post by 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
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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#5
thanks! that would work, but then why would they show both triangles?
(Original post by 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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1 year ago
#6
(Original post by Ljm2001)
thanks! that would work, but then why would they show both triangles?
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.
Last edited by ghostwalker; 1 year ago
1
1 month ago
#7
(Original post by 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.
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
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1 month ago
#8
I acc have a mark scheme answer for this! But it's written down and it's at school
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1 month ago
#9
(Original post by Afrrcyn)
I acc have a mark scheme answer for this! But it's written down and it's at school
nice, mine is getting marked tomorrow so its fine anyway
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1 month ago
#10
Did yours prove 1 + Tan^2 = Sec^2?
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1 month ago
#11
(Original post by 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
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
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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1 month ago
#12
(Original post by 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
and get AD and DC using similar triangles to ABC. A bit longer though and more complex.
I just realised that I used the identity to prove the solution lol, I didn't use the triangle
However I did it again and found a solution lol wait a sec lemme post it
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1 month ago
#13
(Original post by 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
and get AD and DC using similar triangles to ABC. A bit longer though and more complex.
https://photos.app.goo.gl/VUZLP3nMj1siHyWZA
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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1 month ago
#14
(Original post by jacob1140)
https://photos.app.goo.gl/VUZLP3nMj1siHyWZA
heres the answer I got 100% overall marks but also suspected of cheating lol
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?
Last edited by mqb2766; 1 month ago
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1 month ago
#15
(Original post by 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?
Well I never did find an answer to any of the questions so I could't have cheated. Also I'm pretty sure you don't have to explain it with words if you're able to explain it with maths
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1 month ago
#16
(Original post by 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?
Well I never did find an answer to any of the questions so I could't have cheated. Also I'm pretty sure you don't have to explain it with words if you're able to explain it with maths
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1 month ago
#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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