If 8cotθ = 7, then the value of (1 + sinθ) ⁄ cosθ is:

Correct — D. From 8cotθ = 7, cotθ = 7⁄8, so tanθ = 8⁄7.

Take a right triangle with the side opposite θ equal to 8 and the adjacent side equal to 7.

hypotenuse = √(8² + 7²)
= √(64 + 49) = √113

sinθ = 8⁄√113 and cosθ = 7⁄√113.

(1 + sinθ)⁄cosθ = (1 + 8⁄√113) ÷ (7⁄√113)
= ((√113 + 8)⁄√113) × (√113⁄7)
= (8 + √113)⁄7 → option (d)

Why the others are wrong

• (a)
  The divisor has to be the adjacent side, 7, not 8, and the numerator becomes √113 + 8 once 1 and 8⁄√113 are put over one denominator. This choice makes both slips at once.
• (b)
  This is what you get by reading cotθ = 7⁄8 as tanθ = 7⁄8, which swaps the legs: it gives sinθ = 7⁄√113, cosθ = 8⁄√113 and the value (7 + √113)⁄8.
• (c)
  The denominator 7 is right, but the numerator drops the 8. Adding 1 to sinθ = 8⁄√113 gives (√113 + 8)⁄√113, not (√113 + 1)⁄√113.

One trigonometric ratio fixes the shape of the right triangle, and every other ratio can then be read straight off it.

cotθ = adjacent⁄opposite, so 8cotθ = 7 puts the legs in the proportion 7 adjacent to 8 opposite. Pythagoras supplies the hypotenuse, √113.

Because sinθ and cosθ are ratios, the size of the triangle you draw does not matter: legs of 8 and 7 give the same values as legs of 80 and 70.

What is left is fraction work — put 1 + sinθ over the common denominator √113 before you divide.

Key facts

• cotθ = adjacent⁄opposite, so 8cotθ = 7 gives cotθ = 7⁄8 and tanθ = 8⁄7.
• With legs 8 and 7 the hypotenuse is √(64 + 49) = √113.
• √113 stays as it is, because 113 is prime.
• (1 + sinθ)⁄cosθ equals cosθ⁄(1 − sinθ), since (1 + sinθ)(1 − sinθ) = cos²θ.

Common traps

• Reading 8cotθ = 7 as cotθ = 8⁄7. The 8 is a coefficient, so cotθ = 7⁄8.
• Adding 1 to sinθ without a common denominator, which loses the 8 in the numerator.
• Swapping the opposite and adjacent sides, which turns cot into tan and sends 8 to the denominator.

No directly related past PYQ was found.