If 3 sin A + 4 cos A = 5, then the value of tan A is:

Correct — B. Write s for sin A and c for cos A; the second equation you need is s² + c² = 1.

3s + 4c = 5 → s = (5 − 4c)⁄3

Substitute into s² + c² = 1:
(5 − 4c)²⁄9 + c² = 1
25 − 40c + 16c² + 9c² = 9

25c² − 40c + 16 = 0
(5c − 4)² = 0 → c = 4⁄5

s = (5 − 16⁄5)⁄3 = (9⁄5)⁄3 = 3⁄5
tan A = s⁄c = (3⁄5) ÷ (4⁄5) = 3⁄4 → option (b)

Why the others are wrong

• (a)
  5⁄4 is sec A, not tan A. The values found are sin A = 3⁄5 and cos A = 4⁄5, the 3-4-5 triangle, in which 5⁄4 is hypotenuse over adjacent.
• (c)
  3⁄5 is sin A, a value reached on the way. tan A still has to divide it by cos A = 4⁄5, which gives 3⁄4.
• (d)
  4⁄5 is cos A. Reading the coefficients 3 and 4 off the equation as opposite and adjacent lands here without ever using sin² A + cos² A = 1.

One equation in two unknowns looks under-determined, but sin² A + cos² A = 1 is always the second equation available.

Substituting turns 3 sin A + 4 cos A = 5 into a quadratic in cos A, and that quadratic is a perfect square, (5c − 4)² = 0. A single repeated root means A is pinned to one value, not a family of them.

That is by design: a sin A + b cos A can never exceed √(a² + b²), and √(3² + 4²) = 5. The equation sits exactly at the maximum.

Key facts

• sin² A + cos² A = 1 supplies the second equation whenever only a linear relation between sin and cos is given.
• a sin A + b cos A has maximum √(a² + b²) and minimum −√(a² + b²).
• 3 sin A + 4 cos A = 5 sits at that maximum, forcing sin A = 3⁄5 and cos A = 4⁄5.

Common traps

• Reading the coefficients 3 and 4 as opposite and adjacent without checking the identity
• Stopping at cos A = 4⁄5 and answering that
• Expanding (5 − 4c)² as 25 − 16c² and losing the middle term

No directly related past PYQ was found.