The Student Room Group
Reply 1
lil_kk
1) prove that:
secA-cosA=tanAsinA



Here is my solution.
Reply 2
lil_kk
1) (b)solve that giving your answers to in radians to 3d.p

tanAsinA=2 0<A<2pi


Here is my solution. I have not corrected to 3 d.p.

Edit: I've corrected the error which was on picture 232.
lil_kk
2) (a)prove, by counter example, that the statement "sec (A+B)= sec A + sec B, for all A and B" is false

(b)Prove that
tanA+cotA=2cosec2A

2.a.) A = 10, B = 5.
sec(A + B) = sec15 = -1.32 Rads
secA + secB = sec5 + sec10 = 2.33 =/= -1.32 Rads = sec15 = sec(A + B)
-> sec(A + B) =/= secA + sec B for all A and B.

b.) tanA + cotA
= (sinA/cosA) + (cosA/sinA)
= (sin^2A + cos^2A)/(sinAcosA)
= 1/(sinAcosA)
= 1/[(1/2)sin2A]
= 2cosec2A
Reply 4
steve2005
Here is my solution. I have not corrected to 3 d.p.

Edit: I've corrected the error which was on picture 232.


how did you get the second answer??
Reply 5
lil_kk
how did you get the second answer??


A= 1.1437 is the standard answer

but the graph of cosA has a period of 2 (recall the cosine graph)
so there will be two places where cosA = -1 + &#8730;2

one place being A= 1.1437 and to get the other solution, you do:

2pi - 1.1437 = 6.28 - 1.1437 = 5.136

Hope this helps!
Reply 6
Peevee
A= 1.1437 is the standard answer

but the graph of cosA has a period of 2 (recall the cosine graph)
so there will be two places where cosA = -1 + &#8730;2

one place being A= 1.1437 and to get the other solution, you do:

2pi - 1.1437 = 6.28 - 1.1437 = 5.136

Hope this helps!



thanks soo much its made it much clearer