Question

can anyone answer this...
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Answer 1

Solution :

• sinA/sinB = √3/2

=> sin²A/sin²B = 3/4

=> sin²A = 3sin²B/4 --i)

• cosA/cosB = √5/2

=> cos²A/cos²B = 5/4

=> cos²A = 5cos²B/4 --ii)

★ We know that

sin²θ + cos²θ = 1

∴ 3sin²B/4 + 5cos²B/4 = 1

=> 3sin²B + 5(1 - sin²B) = 4

=> 1 = 2sin²B

=> sinB = 1/√2

=> B = 45° , So cosB = cos45° = 1/√2

Now,

=> sinA/sinB = √3/2

=> sinA/(1/√2) = √3/2

=> sinA = √3/2 × √2

=> sinA = √6/2

Also,

=> cosA/cosB = √5/2

=> cosA/(1/√2) = √5/2

=> cosA = √5/2 × √2

=> cosA = √10/2

Now calculate 5tan²A and tan²B

=> 5tan²A + tan²B

=> 5sin²A/cos²A + sin²B/cos²B

=> 5(√6/2/√10/2)² + (1/√2/1/√2)²

=> 5(6/4 × 4/10) + 1

=> 5(3/5) + 1

=> 3 + 1

=> 4

∴ 5tan²A + tan²B = 4

Answer 2

Answer:

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