If 10 sin² θ + 6 cos² θ = 7, 0 < θ < 90°, then find the value of tan θ.

Correct — D. Split the 10 as 6 + 4 so that sin²θ + cos²θ = 1 absorbs one pair.

10 sin²θ + 6 cos²θ = 6(sin²θ + cos²θ) + 4 sin²θ
= 6 + 4 sin²θ = 7
4 sin²θ = 1, so sin²θ = 1⁄4

0 < θ < 90° keeps the positive root, so sin θ = 1⁄2 and θ = 30°
tan 30° = 1⁄√3 → option (d).

Why the others are wrong

• (a)
  Option (a) is 1, i.e. tan 45°. At 45° both squares equal 1⁄2 and the left side becomes 10(1⁄2) + 6(1⁄2) = 8, not the 7 the equation states.
• (b)
  Option (b) is √3, i.e. tan 60°. There sin²θ = 3⁄4 and cos²θ = 1⁄4, giving 10(3⁄4) + 6(1⁄4) = 9. It is also what a candidate who solves for cot θ = √3 reports without inverting.
• (c)
  Option (c) is 1⁄2, which is sin θ, not tan θ — the value you stop at one step early. With sin θ = 1⁄2 and cos θ = √3⁄2, tan θ works out to 1⁄√3.

Any equation shaped a sin²θ + b cos²θ = c yields to one move: write a = b + (a − b) so that b(sin²θ + cos²θ) collapses to the constant b.

With a = 10 and b = 6 the left side becomes 6 + 4 sin²θ, and the equation is linear in sin²θ. The same split run the other way gives 10 − 4 cos²θ = 7, hence cos²θ = 3⁄4 — the same angle by a different route.

The range 0 < θ < 90° is not decoration. sin²θ = 1⁄4 has roots ±1⁄2, and the restriction keeps the positive one and fixes a single angle, 30°.

Key facts

• sin²θ + cos²θ = 1 for every θ, which reduces 10 sin²θ + 6 cos²θ to 6 + 4 sin²θ.
• sin 30° = 1⁄2, cos 30° = √3⁄2 and tan 30° = 1⁄√3 ≈ 0.577.
• The restriction 0 < θ < 90° discards the negative root of sin²θ = 1⁄4.
• The mirror-image split gives 10 − 4 cos²θ, a useful check on the same equation.

Common traps

• Stopping at sin θ = 1⁄2 and reporting that as the value of tan θ.
• Reading 4 sin²θ = 1 as 4 sin θ = 1, which gives sin θ = 1⁄4 and no standard angle.
• Dividing through by sin²θ instead: it does reach cot²θ = 3, but reporting cot θ = √3 as tan θ puts the wrong surd on the sheet.

No directly related past PYQ was found.