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I have been solving these two trigonometric questions for the past 2 days and I still can't derive the answers. I would need someone to help me out with this.
1.Given that cosec A + cot A = 3, evaluate cosec A - cot A and cos A
2.Prove that (1+tan x - sec x)(1 + cot x + cosec x) =2
May I ask whether there is a way to spot the "trigonometric patterns"? I realise sometimes that no matter how much I look at the questions, I still can't figure out a way to get the answers out even though I'm pretty clear with the rules.
I think this topic is similar to that of plane geometry or circle properties. If you see it, bingo, good for you. Otherwise... well.
Thank you anyway!
Edited by anpanman 05 Sep `08, 8:34PM
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Good job on the fact that you have tried it for at least 2 days. Good effort!
I shall attempt :D
1) cosec A + cot A = 3
we know that (cot A)^2 + 1 = (cosec A)^2
Hence, (cosec A)^2 - (cot A)^2 = 1thus, (cosec A + cot A) (cosec A - cot A) = 1
3 (cosec A - cot A) = 1
(cosec A - cot A) = 1/3(cosec A - cot A) = 1/3
(cosec A + cot A) = 3
Summing them, 2 cosec A = 3 1/3
cosec A = 6 2/3 = 20/3
sin A = 0.15Thus, cos A = sqrt (1 - (sin A)^2) = 0.989
2) Prove that (1+tan x - sec x)(1 + cot x + cosec x) =2
expand
LHS= 1 + cot x + cosec x + tan x + 1 + tan x cosec x - sec x - sec x cot x - sec x cosec x
We can calculate that
tan x cosec x = sec x (since tan x = sin x / cos x)
sec x cot x = cosec xso the above is
LHS = 1 + cot x + cosec x + tan x + 1 + sec x - sec x - cosec x - sec x cosec x
LHS = 2 + cot x + tan x - sec x cosec x
LHS = 2 + cos x / sin x + sin x / cos x - 1 / (sin x cos x)
LHS = 2 + [{cos x}^2 + {sin x}^2 - 1] / (sin x cos x)
LHS = 2 (proved)Regards,
Eagle (Strategic Tuition)
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Originally posted by eagle:
Good job on the fact that you have tried it for at least 2 days. Good effort!
I shall attempt :D
1) cosec A + cot A = 3
we know that (cot A)^2 + 1 = (cosec A)^2
Hence, (cosec A)^2 - (cot A)^2 = 1thus, (cosec A + cot A) (cosec A - cot A) = 1
3 (cosec A - cot A) = 1
(cosec A - cot A) = 1/3(cosec A - cot A) = 1/3
(cosec A + cot A) = 3
Summing them, 2 cosec A = 3 1/3
cosec A = 6 2/3 = 20/3
sin A = 0.15Thus, cos A = sqrt (1 - (sin A)^2) = 0.989
2) Prove that (1+tan x - sec x)(1 + cot x + cosec x) =2
expand
LHS= 1 + cot x + cosec x + tan x + 1 + tan x cosec x - sec x - sec x cot x - sec x cosec x
We can calculate that
tan x cosec x = sec x (since tan x = sin x / cos x)
sec x cot x = cosec xso the above is
LHS = 1 + cot x + cosec x + tan x + 1 + sec x - sec x - cosec x - sec x cosec x
LHS = 2 + cot x + tan x - sec x cosec x
LHS = 2 + cos x / sin x + sin x / cos x - 1 / (sin x cos x)
LHS = 2 + [{cos x}^2 + {sin x}^2 - 1] / (sin x cos x)
LHS = 2 (proved)Regards,
Eagle (Strategic Tuition)
Wow, splendid! Thank you!P.S. you made a minor mistake in the first question.
Summing them, 2 cosec A = 3 1/3
cosec A = 6 2/3 = 20/3It should be 5/3 instead of 20/3...
Thanks a lot! ;p
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