Q.287. Infinite trigonometric geometric sequence satisfying a quadratic equation

Question 287.
Expression ( sin²x + sin⁴x + sin⁶x + ........ + ∞ ) log 3 satisfies the expression x² - 28 x + 27 = 0, then Find the value of  cosx / ( sinx + cosx ).  [x belongs to First quadrant.]

Answer 287.
x^2 - 28x + 27 =0
=> (x - 27)(x - 1) = 0
=> x = 27 or x = 1 ... ( 1 )
 sin^2 x +sin^4 x + sin^ 6x + ...to ∞
= sin^2 x /(1 - sin^2 x)
= tan^2 x

Plugging tan^2 x * log3 in place of x in ( 1 ),
log 3^(tan^2 x) = log 27 or log 1
=> 3^(tan^2 x) = 3^3 or 1
=>tan^2 x = 3 or 0
=> x = kπ ± π/3 or kπ

For x belonging to the first quadrant, x = π/3
=> cosx / (sinx + cosx)
= cos(π/3) / [sin(π/3) + cos(π/3)]
= (1/2) /[(√3/2+1/2)]
= 1/(√3 + 1)
= (√3 - 1)/2.

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