Find the value of a for which

4cosec² ( π (a + x)) +a² - 4a = 0 has a real solution.

Dear Nirabhra,

Solution:- 4cosec²(n(a+x)) + a² - 4a =0

[4cosec²(n(a+x)) -4] + (a² - 4a + 4) =0

4[cosec²(n(a+x)) -1] + (a-2)² =0

[2cot(n(a+x))]² + (a-2)² =0 which means a =2 and cot(n(a+x)) =0

Ans: a= 2

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Dear Nirabhra,

Solution:- 4cosec²(n(a+x)) + a² - 4a =0

[4cosec²(n(a+x)) -4] + (a² - 4a + 4) =0

4[cosec²(n(a+x)) -1] + (a-2)² =0

[2cot(n(a+x))]² + (a-2)² =0 which means a =2 and cot(n(a+x)) =0

4 cosec ² (π (a+x) ) =- a² + 4a 

L .H .S is always greater than 4 so for real solution

 -a² + 4a ≥4

or (a-2)²≤0

so only a=2 is possible