If 3sin²θ + 5cos²θ = 4 and θ is an acute angle, then the value of tanθ is:

Correct — C. Split the 5cos²θ so that sin²θ + cos²θ = 1 absorbs part of it.

3sin²θ + 5cos²θ = 3(sin²θ + cos²θ) + 2cos²θ
= 3 + 2cos²θ

Set that equal to 4:
3 + 2cos²θ = 4 → cos²θ = 1⁄2
cos θ = 1⁄√2, and θ is acute, so θ = 45°

tan 45° = 1 → option (c)

Why the others are wrong

• (a)
  1⁄√3 is tan 30°, which would need cos²θ = 3⁄4. Put that back into the identity and 3 + 2(3⁄4) = 4.5, not the 4 the stem states.
• (b)
  1⁄2 is cos²θ, lifted out of the working one line too early. It is not the tangent of any angle this equation produces, and cos²θ still has to be turned into tan θ.
• (d)
  1⁄√2 is cos θ itself. The working genuinely reaches cos θ = 1⁄√2 and then stops, instead of naming the angle, 45°, and taking its tangent.

Every question of this shape rests on sin²θ + cos²θ = 1. Rewrite the sin²–cos² combination as a multiple of that identity plus one leftover term, and an unknown disappears.

Here 3sin²θ + 5cos²θ becomes 3 + 2cos²θ, so the equation pins cos²θ and nothing else is needed — you never solve for sin θ and cos θ separately.

The acute condition is doing real work: cos²θ = 1⁄2 admits 45° and 135°, and only 45° is acute, so tan θ is +1 and not −1.

Key facts

• sin²θ + cos²θ = 1 for every value of θ.
• 3sin²θ + 5cos²θ simplifies to 3 + 2cos²θ.
• cos 45° = sin 45° = 1⁄√2, and tan 45° = 1.

Common traps

• Answering with cos θ once cos²θ has been found, instead of converting to tan θ.
• Ignoring the acute rider and keeping the 135° root, which flips the sign.
• Trying to find sin θ and cos θ individually when only their squares are constrained.

No directly related past PYQ was found.