Evaluate the given expression.
- (a)0
- (b)2
- (c)3
- (d)5
Answer
Why
Correct — D. Two Pythagorean identities clear both denominators.
1 + cot²θ = cosec²θ, so 5 ⁄ (1 + cot²θ) = 5 sin²θ
1 + tan²θ = sec²θ, so 3 ⁄ (1 + tan²θ) = 3 cos²θ
Expression = 5 sin²θ + 3 cos²θ + 2 cos²θ
= 5 sin²θ + 5 cos²θ
= 5 (sin²θ + cos²θ) = 5 × 1 = 5 → option (d).
Why the others are wrong
- (a)0 — 0 is impossible for any θ. Every term reduces to a square times a positive number, and sin²θ and cos²θ are never both zero, so the sum cannot vanish.
- (b)2 — 2 copies the coefficient of the trailing cos² term. That term is 2 cos²θ, which reaches 2 only at cos θ = ±1, and the two fractions do not disappear.
- (c)3 — 3 copies the numerator of the second fraction. Its value is 3 cos²θ, which equals 3 only when cos θ = ±1, and no value of θ is fixed in the question.
Concept
Two identities do all the work: 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ. Both come from dividing sin²θ + cos²θ = 1 by cos²θ and by sin²θ.
The reciprocals are the useful form here, since the identities sit in denominators: 1 ⁄ (1 + tan²θ) = cos²θ and 1 ⁄ (1 + cot²θ) = sin²θ.
The expression is then designed so the coefficients match. Five sin²θ meets five cos²θ once the trailing 2 cos²θ is added to the 3 cos²θ, and sin²θ + cos²θ = 1 collapses everything to a number that does not depend on θ.
Nothing in the question fixes θ, which is itself the clue that the expression must be constant. Strictly the working holds wherever tan θ and cot θ are both defined, so not at 0° or 90°.
Key facts
- sin²θ + cos²θ = 1 for every θ.
- 1 + tan²θ = sec²θ, so 1 ⁄ (1 + tan²θ) = cos²θ.
- 1 + cot²θ = cosec²θ, so 1 ⁄ (1 + cot²θ) = sin²θ.
- Substituting θ = 45° gives 5(½) + 3(½) + 2(½) = 5, the same constant.
Study next
Common traps
- Swapping the two identities and writing 1 + cot²θ as sec²θ.
- Leaving the answer as 5 sin²θ + 5 cos²θ without collapsing it to 5.
- Substituting a convenient angle, which works here but hides the identity and fails when the expression is not constant.
The design is that the coefficients are chosen so the sin² and cos² terms end up equal, leaving the identity to finish the job.
A different route to a θ-free value is asked on 25 Sep 2024, 09:00, Quant Q.1, which multiplies (cosec θ − sin θ)(sec θ − cos θ)(tan θ + cot θ).
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