Question

If sin A/3 = sin B/4 =1/5 and A,B are angles in the second quadrant then prove that 4cosA + 3cosB = -5.​

Answer 1

Answer:

Given:

• sin A/3 = sin B/4 = 1/5 and A, B are angles in the second quadrant.

To prove:

• 4cos A + 3cos B = - 5

Solution:

=> sin A/3 = 1/5

=> sin A = 3/5

On squaring both sides

=> sin²A = 9/25

By using identity, cos²Ø = 1 - sin²Ø

=> cos²A = 1 - sin²A

=> cos²A = 1 - 9/25

=> cos²A = 16/25

Taking square root on both sides

=> cos A = -4/5...(1) [angle is in quadrant II]

Also,

=> sin B/4 = 1/5

=> sin B = 4/5

On squaring both sides

=> sin²B = 16/25

By using identity, cos²Ø = 1 - sin²Ø

=> cos²B = 1 - sin²B

=> cos²B = 1 - 16/25

=> cos²B = 9/25

On taking square root on both sides

=> cos B = -3/5...(2) [angle is in quadrant II]

Now, 4cos A + 3cos B = 4 (-4/5) + 3 (-3/5)

...from (1) and (2)

=> 4cos A + 3cos B = (- 16 - 9)/5

=> 4cos A + 3cos B = -25/5

=> 4cos A + 3cos B = - 5

Hence, proved.

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Note:

• All angles are positive in first quadrant.

• sinØ and cosecØ are only positive in second quadrant.

• tanØ and cotØ are only positive in third quadrant.

• cosØ and secØ are only positive in forth quadrant.