d:[["$","title","0",{"children":"IIM Kozhikode - If 1 + ^2 = 3 , then can take values | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined Looking at this equation, I need to find all possib<= values of . Let me work through this using the easiest approach.

Given: 1 + ^2 = 3 I'll use the identity: ^2 + ^2 = 1, which means ^2 = 1 - ^2

Substituting into the original equation: 1 + (1 - ^2) = 3 2 - ^2 = 3

Dividing everything by ^2 to >=t : 2 - ^2^2 = 3 ^2 2^2 - 1 = 3

Using the identities 1^2 = 1 + ^2 and = : 2(1 + ^2) - 1 = 3 2 + 2^2 - 1 = 3 1 + 2^2 = 3

Rearranging into standard quadratic form: 2^2 - 3 + 1 = 0

Let t = , so: 2t^2 - 3t + 1 = 0 Using the quadratic formula: t = 3 \\± sqrt(9 - 8)4 = 3 \\± 14 Therefore: t = 44 = 1 or t = 24 = 12

Verifying our solutions: For = 1: This means = 45°, so = = 1sqrt(2) LHS: 1 + 12 = 32 RHS: 3 1sqrt(2) 1sqrt(2) = 32 ✓ For = 12: Using = 1sqrt(5), = 2sqrt(5) Both sides equal 65 ✓

can take the values 12 and 1"}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"IIM-K (BMSAT),QA,Geometry,Trigonometry"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"property":"og:title","content":"IIM Kozhikode - If 1 + ^2 = 3 , then can take values | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined Looking at this equation, I need to find all possib<= values of . Let me work through this using the easiest approach.

Given: 1 + ^2 = 3 I'll use the identity: ^2 + ^2 = 1, which means ^2 = 1 - ^2

Substituting into the original equation: 1 + (1 - ^2) = 3 2 - ^2 = 3

Dividing everything by ^2 to >=t : 2 - ^2^2 = 3 ^2 2^2 - 1 = 3

Using the identities 1^2 = 1 + ^2 and = : 2(1 + ^2) - 1 = 3 2 + 2^2 - 1 = 3 1 + 2^2 = 3

Rearranging into standard quadratic form: 2^2 - 3 + 1 = 0

Let t = , so: 2t^2 - 3t + 1 = 0 Using the quadratic formula: t = 3 \\± sqrt(9 - 8)4 = 3 \\± 14 Therefore: t = 44 = 1 or t = 24 = 12

Verifying our solutions: For = 1: This means = 45°, so = = 1sqrt(2) LHS: 1 + 12 = 32 RHS: 3 1sqrt(2) 1sqrt(2) = 32 ✓ For = 12: Using = 1sqrt(5), = 2sqrt(5) Both sides equal 65 ✓

can take the values 12 and 1"}],["$","meta","9",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/iimk-bmsat/2025/QA/11"}],["$","meta","10",{"property":"og:site_name","content":"AfterBoards | IIM Kozhikode Preparation"}],["$","meta","11",{"property":"og:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","meta","12",{"property":"og:type","content":"website"}],["$","meta","13",{"name":"twitter:card","content":"summary_large_image"}],["$","meta","14",{"name":"twitter:site","content":"afterboards.in"}],["$","meta","15",{"name":"twitter:title","content":"IIM Kozhikode - If 1 + ^2 = 3 , then can take values | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined Looking at this equation, I need to find all possib<= values of . Let me work through this using the easiest approach.

Given: 1 + ^2 = 3 I'll use the identity: ^2 + ^2 = 1, which means ^2 = 1 - ^2

Substituting into the original equation: 1 + (1 - ^2) = 3 2 - ^2 = 3

Dividing everything by ^2 to >=t : 2 - ^2^2 = 3 ^2 2^2 - 1 = 3

Using the identities 1^2 = 1 + ^2 and = : 2(1 + ^2) - 1 = 3 2 + 2^2 - 1 = 3 1 + 2^2 = 3

Rearranging into standard quadratic form: 2^2 - 3 + 1 = 0

Let t = , so: 2t^2 - 3t + 1 = 0 Using the quadratic formula: t = 3 \\± sqrt(9 - 8)4 = 3 \\± 14 Therefore: t = 44 = 1 or t = 24 = 12

Verifying our solutions: For = 1: This means = 45°, so = = 1sqrt(2) LHS: 1 + 12 = 32 RHS: 3 1sqrt(2) 1sqrt(2) = 32 ✓ For = 12: Using = 1sqrt(5), = 2sqrt(5) Both sides equal 65 ✓

can take the values 12 and 1"}],"$L17","$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f","$L20"]

IIM Kozhikode (QA) PYQs

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