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Solve for X . tanx + secx = 3 - ValuSoft Bible Collection (10281) for PC

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Hint: split it up into 
int sec^2(x) dx + int e^sin(x) cos(x) dx
The first one is standard, and the second one is a straightforward u-substitution.

Posted on Jul 20, 2009

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Tanx+coty/tanx coty


I going to use the following trig identies:
cotx = 1/tanx, and tanx = 1/cotx

Therefore:
(tanx+coty)/(tanx*coty) = (tanx + 1/tany)/(tanx/tany)
multiplying the top and bottom by tany yields:
(tany*tanx + 1) / tanx
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tany + 1/tanx
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(1+cotx-cosecx)(1+tanx+secx)=2


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 = 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 = 6 2/3 = 5/3
sin A = 0.15
Thus, 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 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 + [{cos x}^2 + {sin x}^2 - 1] / (sin x cos x)
LHS = 2 (proved)

May 12, 2009 | ValuSoft Bible Collection (10281) for PC

3 Answers

Sinx+cosx=5, tanx+secx=3 then sinx=?


tanx= sinx/cosx,
secx=1/cosx
1+sinx/cosx=3
1+sinx=3cosx---------->1
cosx=5-sinx------------->2

sub 2 in 1
1+sinx=3(5-sinx)
1+sinx=15-3sinx
1+4sinx=15
4sinx=14
sinx=14/4...
this is the solution..

Jul 14, 2008 | ValuSoft Bible Collection (10281) for PC

2 Answers

Tanx a lot dude!!!


No problem bro. Anytime.

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