If tan 2 ⁡ x + y + cot 2 ⁡ x + y = 1 - 2 x - x 2 , then w h e r e , n ∈ Z

Question

If tan 2 ⁡ x + y + cot 2 ⁡ x + y = 1 - 2 x - x 2 , then w h e r e , n ∈ Z, x = 1 , y = n π ± π 4 - 1, x = - 1 , y = n π ± π 4 + 1, x = - 1 , y = n π 2 ± π 4 - 1, x = + 1 , y = n π 2 ± π 4 - 1

MARKS App · Accepted Answer

Left hand side is greater than or equal to 2 . ⇒ 1 - 2 x - x 2 ≥ 2 ⇒ x + 1 2 ≤ 0 ⇒ x = - 1 Hence, the given equation is tan 2 ⁡ y - 1 + cot 2 ⁡ y - 1 = 2 ⇒ tan ⁡ y - 1 = ± 1 ⇒ y - 1 = nπ ± π 4 ⇒ y = nπ ± π 4 + 1 So, x = - 1 and y = n π ± π 4 + 1

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