Determine the integral values of 'k' for which the system, (tan –1 x) 2 + (cos –1 y) 2 = π 2 k and
tan –1 x + cos –1 y = 
possess solution and find all the solutions.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
k = 1, x = tan (1 – 
) 
, y = cos ( 
+ 1) 
Sol. 
⇒ (tan –1 x) 2 + (cos –1 x) 2 ≤ 
But (tan –1 x) 2 + (cos –1 x) 2 = π 2 k hence k π 2 ≤ 
, k ≤ 
.......(i)
Now put tan –1 x = 
– cos –1 y 
+ (cos –1 y) 2 = π 2 k (where cos –1 y = t)
2t 2 – π t + 
= 0
For real roots, D ≥ 0
π 2 – 8 
≥ 0 ⇒ 1 – 2 + 8k ≥ 0 , k ≥ 
....(ii)
From (i) and (ii), k = 1
With k = 1, t = 
= 
= (1 ± 
) 
.
or cos –1 y = ( 
+ 1) 
(as 0 ≤ cos –1 y ≤ π ) ∴ y = cos ( 
+ 1) 
∴ tan –1 x = 
– ( 
+ 1) 
= 
[(1 – 
)] ⇒ x = tan (1 – 
) 
.
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