The Student Room Group

Trigonometry

Figure 1 shows the right-angled triangles ABC, ABD and BDC, with AB=1 and ÐBAD=q .
Prove that 1+tan^2q =sec^2q .
(edited 4 years ago)
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?
Reply 2
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?
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 cos2θ+sin2θ=1\cos^2\theta+\sin^2\theta =1 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.
Reply 4
thanks! that would work, but then why would they show both triangles?
Original post by ghostwalker
I guess you're refering to the cos2θ+sin2θ=1\cos^2\theta+\sin^2\theta =1 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.
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.
(edited 4 years ago)
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
https://photos.google.com/share/AF1QipNzjBjl_ULCXopdE0a1Qe9CL9-dVrl4vmoqvlDUTHMDd3KZZ5jdUkj2ihvdhFDHdw?key=VGdORU4ybEFRNUQzaXEycXhDVTVhckVZWEpMb1ZR
I acc have a mark scheme answer for this! But it's written down and it's at school :frown:
Original post by Afrrcyn
I acc have a mark scheme answer for this! But it's written down and it's at school :frown:

nice, mine is getting marked tomorrow so its fine anyway
Did yours prove 1 + Tan^2 = Sec^2?
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
https://photos.google.com/share/AF1QipNzjBjl_ULCXopdE0a1Qe9CL9-dVrl4vmoqvlDUTHMDd3KZZ5jdUkj2ihvdhFDHdw?key=VGdORU4ybEFRNUQzaXEycXhDVTVhckVZWEpMb1ZR

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.
(edited 3 years ago)
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
AC = AD + DC
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
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
AC = AD + DC
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
(edited 3 years ago)
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?
(edited 3 years ago)
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
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
Reply 16
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
Reply 17
could you upload the picture again it bc it’s not there
Reply 18
Original post by username5424512
https://photos.app.goo.gl/VUZLP3nMj1siHyWZA
heres the answer I got 100% overall marks but also suspected of cheating lol

can you upload the picture again bc it’s not there
Original post by kenza.o
can you upload the picture again bc it’s not there

Why not try and do it and upload your attempt if you have problems.

Quick Reply

Latest