Math, asked by sandeepchaudhary9860, 16 days ago

Solve this 2sin(A+15°)=√3

Answers

Answered by 31aliahmedzahidshaik

0

Answer:

Given:

⇒ 2sin(3x -15)° = 1

Calculation:

⇒ sin(3x - 15)° = 1/2

⇒ (3x -15)° = sin-1(1/2)

But, 0° < (3x - 15) < 90° then,

⇒ (3x - 15)° = 30°

⇒ x = 15°

Then,

⇒ ? = cos2(2x +15)° + cot2 (x +15)°

⇒ ? = cos2(45)° + cot2(30)°

⇒ ? = 1/2 + 3

⇒ ? = 7/2

∴ cos2(2x +15)° + cot2 (x +15)° = 7/2

Answered by priyaranjanmanasingh

0

Answer:

Given:

⇒ 2sin(3x -15)° = 1

Calculation:

⇒ sin(3x - 15)° = 1/2

⇒ (3x -15)° = sin-1(1/2)

But, 0° < (3x - 15) < 90° then,

⇒ (3x - 15)° = 30°

⇒ x = 15°

Then,

⇒ ? = cos2(2x +15)° + cot2 (x +15)°

⇒ ? = cos2(45)° + cot2(30)°

⇒ ? = 1/2 + 3

⇒ ? = 7/2

∴ cos2(2x +15)° + cot2 (x +15)° = 7/2

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