A mind boggling maths question?

Question

(1+(1+x)1/2)tanA=1-{(1-x)1/2}
find the value of x in terms of A.,give the entire solution...
NOTE:1/2 is in power of (1+x) and (1-x)

Answer 1

because we are taking (1+x)^(1/2) and (1-x)^(1/2) we can put

x= sin 2t
(1+x)^1/2 = (1- sin 2t)^(1/2) = (sin^2 t+cos^2 t + 2sint cost)^(1/2) = sin t + cos t
(1-x)^(1/2) = sint - cos t

so tan A ( 1 + sint + cos t) = 1 + sint - cos t

tan A = (1+ sint + cos t)/(1+ sint - cos t) ....1

Simplify this to get

tan A = (1+cos t)/ sin t

tan A = sin t /(1- cos t)

cot A = (1- cos t)/ sin t

add (Tan A + cot A) = 2/sin t

subtract Tan A - cot A = 2 cos t/sin t

so sin t = 2/(Tan A + cot A)

cos t = (tan A- cot A) sin t /2 = (tan A - cot A)/(Tan A + cot A)
multiply and get x after simplifying
x = 2 sin t cos t = 4(tan A-cot A)/(tan A + cot A)^2

Answer 2

we have

tan A = (1- (1-x)^1/2 )
@@ ---------------
@@@ (1+ (1-x)^1/2 )





and also

cot A = 1/tan A

Adding 1 on both sides

1+ cot A = 2 / {1 + (! + x)^1/2 }

2 / (1+cot A ) = 1 + (1+ x)^1/2

1+x = [ { 1-cotA}^1/2 / {1+cotA}^1/2 ]

Dividing the numerator and denominator on the r.h.s by tanA and solving the equation

x = { tanA -1 }^2 - 1
@@ -----------------
@@ {tanA +1)^2

This should Explain u

Answer 3

Hint.

Any problem with exponents can be made relatively easy by taking Ln on both side. ( ln or Lg)

Try it.

Answer 4

see questions like this is why i hated math im never going to have to use this in real life i dont think anyone will unless they are a scientist or something like that .....any one agree?

Answer 5

why? why did you have to go an hurt my head like that? and make me feel very very stupid!

Answer 6

x=logA-A