Turn on thread page Beta
    • Thread Starter
    Offline

    1
    ReputationRep:
    Hello, TSR!

    Could any A2 Mathematicians please help me to prove the following trig identity:

    (1-cosx)/sinx = 1/(cosecx + cotx)

    Many thanks!
    Offline

    1
    ReputationRep:
    What can you replace cosec(x) and cot(x) with?

    Then simplify.
    Offline

    3
    ReputationRep:
    Split the LHS into two fractions: 1/sinx - cosx/sinx

    cosecx - cotx

    multiply by [ (cosecx + cotx) / (cosecx + cotx) ]

    [ (cosecx -cotx) . (cosecx + cotx) ] / (cosecx + cotx)

    (cosec^2x - cot^2x) / (cosecx + cotx)

    1 / (cosecx + cotx)

    L.H.S = R.H.S
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by Drackore)
    Hello, TSR!

    Could any A2 Mathematicians please help me to prove the following trig identity:

    (1-cosx)/sinx = 1/(cosecx + cotx)

    Many thanks!
    There will be a few ways to do this one. I would multiply the top and bottom of the RHS by cosec \ x - cot \ x.
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by AmarPatel98)
    Split the LHS into two fractions: 1/sinx - cosx/sinx

    cosecx - cotx

    multiply by [ (cosecx + cotx) / (cosecx + cotx) ]

    [ (cosecx -cotx) . (cosecx + cotx) ] / (cosecx + cotx)

    (cosec^2x - cot^2x) / (cosecx + cotx)

    1 / (cosecx + cotx)

    L.H.S = R.H.S
    Please don't post full solutions. It's against the rules of this forum.
    • Thread Starter
    Offline

    1
    ReputationRep:
    Thanks very much for the replies, guys! Ultimately it was Amar's method that I used, as I'd already made a start with that one but didn't know where to go from from cosecx - cotx
    Offline

    3
    ReputationRep:
    (Original post by notnek)
    Please don't post full solutions. It's against the rules of this forum.
    My apologies... I was unaware of this
    Offline

    0
    ReputationRep:
    Attachment 461189
    #1


    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    I'm going to oversimplify; realistically you would do most of this in your head.You could have alternatively split up (1-cosx)/sinx in terms of cosecx & secx then multiply top and bottom by cosecx + cotx .
    cosecx + cotx=(1/sinx) +(cosx/sinx) =(1+cosx)/sinx

    therefore 1/(cosecx + cotx)= sinx/(1+cosx)
    = sinx(1-cosx)/(1+cosx)(1-cosx)
    = sinx(1-cosx)/1-(cosx)^2
    1-(cosx)^2=1-[1-(sinx)^2)] =(sinx)^2

    therefore 1/(cosecx + cotx) =sinx(1-cosx)/(sinx)^2
    = (1-cosx)/sinx
    L.H.S = R.H.S
    Offline

    6
    ReputationRep:
    (Original post by Nick Moore)
    I'm going to oversimplify; realistically you would do most of this in your head.You could have alternatively split up (1-cosx)/sinx in terms of cosecx & secx then multiply top and bottom by cosecx + cotx .
    cosecx + cotx=(1/sinx) +(cosx/sinx) =(1+cosx)/sinx

    therefore 1/(cosecx + cotx)= sinx/(1+cosx)
    = sinx(1-cosx)/(1+cosx)(1-cosx)
    = sinx(1-cosx)/1-(cosx)^2
    1-(cosx)^2=1-[1-(sinx)^2)] =(sinx)^2

    therefore 1/(cosecx + cotx) =sinx(1-cosx)/(sinx)^2
    = (1-cosx)/sinx
    L.H.S = R.H.S
    Hey - is there any chance you could provide me with a pointer for this question?

    Prove this trig identity:

    (\sin \theta - \mathrm{cosec } \theta)^2 \equiv \sin^2 \theta + \mathrm{cot}^2 \theta -1

    I started with the left side, and got up to

     \sin^2 \theta - \frac{2 \sin \theta}{\sin \theta} + \frac{1}{\sin^2 \theta}

    But not really sure where to go from here to end up with

     \sin^2 \theta + \mathrm{cot}^2 \theta -1
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (Original post by Funky_Giraffe)
    Hey - is there any chance you could provide me with a pointer for this question?

    Prove this trig identity:

    (\sin \theta - \mathrm{cosec } \theta)^2 \equiv \sin^2 \theta + \mathrm{cot}^2 \theta -1

    I started with the left side, and got up to

     \sin^2 \theta - \frac{2 \sin \theta}{\sin \theta} + \frac{1}{\sin^2 \theta}

    But not really sure where to go from here to end up with

     \sin^2 \theta + \mathrm{cot}^2 \theta -1
    SImplifying the line you're up to:

    \sin^2 \theta -2 + cosec^2 \theta

    Which I'll now write in another way:

    \sin^2 \theta+ cosec^2\theta - 1 - 1


    Does that help?
    Offline

    6
    ReputationRep:
    (Original post by notnek)
    SImplifying the line you're up to:

    \sin^2 \theta -2 + cosec^2 \theta

    Which I'll now write in another way:

    \sin^2 \theta+ cosec^2\theta - 1 - 1


    Does that help?
    Thanks very much!
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: September 19, 2015
Poll
Cats or dogs?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.