Question

1 . sin4A = cos(A-20°)

2. Sin38°/Cos52° = sin38° - sin(90°-52°)

3. cos80°/sin10° + cos59° cosec401°

4. sinø + cos ø = √2 . cos ( 90°-ø)

5. sec70° . sin20° - cos20° . cosec70°

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Answer 1

Solutions

Q.1. sin4A = cos(A-20°)

=> cos ( 90°-4A) = cos(A-20°)

•

• • [ sin Ø = cos(90°- Ø) ]

=> 90° - 4A = A - 20°

=> 5A = 110°

=> A = 110°/5 = 22°

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Q . 2. sin38° - cos52° = sin38° - sin(90°-52°)

=> sin 38° - sin 38°

=> 0 [ As, ( cos A = sin(90°-A) ) ]

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Q.3. cos80°/sin10° + cos59° cosec31°

• using sin(90°-Ø ) = cosØ , we get ;

=> sin10° = sin(90°-80°) = cos 80° and cos59° = cos(90°-31°) = sin 31 °

• Thus , cos80°/sin80° + sin31° cosec31°

=> 1 + 1/cosec31° × cosec 31°

=> 1+1 = 2

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Q.4. sin Ø + cosØ = √2 . cos ( 90°- Ø )

=> sin Ø + cos Ø = √2 sin Ø

=> cos Ø = √2 sin Ø - sin Ø

=> cos Ø = (√2-1)sinØ

=> cos Ø / sin Ø = √2-1

=> cot Ø = √2 - 1

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Q.5. sec70° . sin20° - cos20° . cosec70°

=> cosec ( 90° - 70° ) . sin20° - cos30° . sec(90°-70°)

=> cosec 20° × 1/cosec20° - 1/sec20° × sec20°

=> 1 - 1 = 0

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