If 7 tan θ = 3, and θ is an acute angle, then (5 sin θ − cos θ) ⁄ (5 sin θ + 2 cos θ) is equal to:

Correct — C. Every term is degree one in sin θ and cos θ, so divide top and bottom by cos θ and only tan θ survives.

7 tan θ = 3 → tan θ = 3⁄7

(5 sin θ − cos θ)/(5 sin θ + 2 cos θ)
= (5 tan θ − 1)/(5 tan θ + 2)

5 tan θ = 15⁄7
Numerator = 15⁄7 − 7⁄7 = 8⁄7
Denominator = 15⁄7 + 14⁄7 = 29⁄7

(8⁄7) ÷ (29⁄7) = 8⁄29 → option (c).

Why the others are wrong

• (a)
  The denominator 29 is right, so the slip is upstairs. 5 tan θ − 1 = 15⁄7 − 7⁄7 = 8⁄7, and no step in the working produces 11⁄7.
• (b)
  The 1 converted wrongly. Over a common denominator of 7 the standalone 1 is 7⁄7, so 15⁄7 − 7⁄7 = 8⁄7. Subtracting 8⁄7 instead would leave 7⁄7.
• (d)
  Would need 5 tan θ = 16⁄7. With tan θ = 3⁄7 the product is 15⁄7, so the numerator lands on 8⁄7, not 9⁄7.

When every term of a fraction has the same degree in sin θ and cos θ, dividing numerator and denominator by cos θ to that degree turns the whole thing into an expression in tan θ alone.

Here 5 sin θ − cos θ and 5 sin θ + 2 cos θ are both degree one, so a single division gives (5 tan θ − 1)/(5 tan θ + 2).

That is why the stem hands you 7 tan θ = 3 and nothing else — the angle θ itself never has to be found.

Key facts

• Dividing a degree-one homogeneous expression in sin θ and cos θ by cos θ leaves an expression in tan θ.
• 7 tan θ = 3 gives tan θ = 3⁄7, so 5 tan θ = 15⁄7.
• (5 tan θ − 1)/(5 tan θ + 2) = (8⁄7)/(29⁄7) = 8⁄29.

Common traps

• Converting the standalone 1 to 1⁄7 instead of 7⁄7 when subtracting from 15⁄7.
• Inverting the given ratio and working with tan θ = 7⁄3.
• Dividing by sin θ instead of cos θ and being left with cot θ terms.

No directly related past PYQ was found.