Standard trigo question from http://www.sgforums.com/forums/2297/topics/329607
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
Answer:
Q 1)
cosec A + cot A = 3
we know that cot² A + 1 = cosec² A
Hence, cosec² A - cot² A = 1
thus, (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 = 5/3
sin A = 0.6
Thus, cos A = sqrt (1 - sin² A) = 0.8
Q 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 x
so 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 + [cos² x + sin² x - 1] / (sin x cos x)
LHS = 2 (proved)
