JEE Main 2025 — 23 January, Shift 1. Previous Year Question.
Problem
The value of $(\sin 70^\circ)(\cot 10^\circ \cot 70^\circ – 1)$ is equal to:
Key insight. Writing $\cot 10^\circ \cot 70^\circ – 1$ as a single fraction over $\sin 10^\circ \sin 70^\circ$ turns the numerator into exactly the expansion of $\cos(A+B)$ — and $10^\circ + 70^\circ = 80^\circ$ is the complement of $10^\circ$, which is what makes everything collapse.
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Approach
Whenever cotangents appear multiplied together with a $-1$ alongside them, it’s worth converting everything to sines and cosines and combining over a common denominator — this often exposes a $\cos(A+B)$ or $\cos(A-B)$ pattern. Here, doing that conversion and recognising $\cos 80^\circ$ as $\sin 10^\circ$ (complementary angles) is enough to collapse the whole expression to a simple ratio.
Solution
Step 1 — Convert cotangents to sines and cosines
$$\cot 10^\circ \cot 70^\circ – 1 = \frac{\cos 10^\circ \cos 70^\circ}{\sin 10^\circ \sin 70^\circ} – 1 = \frac{\cos 10^\circ\cos 70^\circ – \sin 10^\circ \sin 70^\circ}{\sin 10^\circ \sin 70^\circ}$$
Step 2 — Recognise the numerator as $\cos(A+B)$
Using $\cos A\cos B – \sin A \sin B = \cos(A+B)$:
$$\cos 10^\circ\cos 70^\circ – \sin 10^\circ\sin 70^\circ = \cos(10^\circ+70^\circ) = \cos 80^\circ$$
So:
$$\cot 10^\circ\cot 70^\circ – 1 = \frac{\cos 80^\circ}{\sin 10^\circ \sin 70^\circ}$$
Step 3 — Multiply by $\sin 70^\circ$ and simplify
$$\sin 70^\circ \times \frac{\cos 80^\circ}{\sin 10^\circ\sin 70^\circ} = \frac{\cos 80^\circ}{\sin 10^\circ}$$
Step 4 — Use the complementary angle identity
Since $\cos 80^\circ = \cos(90^\circ – 10^\circ) = \sin 10^\circ$:
$$\frac{\cos 80^\circ}{\sin 10^\circ} = \frac{\sin 10^\circ}{\sin 10^\circ} = 1$$
Answer
$$1$$
Common mistakes
- Trying to evaluate $\cot 10^\circ$ and $\cot 70^\circ$ numerically rather than combining them symbolically — this makes the complementary-angle cancellation invisible and turns a two-line simplification into a messy decimal calculation.
- Missing that $10^\circ + 70^\circ = 80^\circ$ is complementary to $10^\circ$ (since $80^\circ + 10^\circ = 90^\circ$) — this is the specific fact that makes $\cos 80^\circ = \sin 10^\circ$, and without it the final cancellation doesn’t happen.
Practise next
- Evaluate $(\sin 50^\circ)(\cot 20^\circ\cot 50^\circ – 1)$ using the same cos(A+B) approach, checking which angle pair is complementary.
Show answer
$1$. Combine the bracket over a common denominator:
$$\cot20^\circ\cot50^\circ-1=\frac{\cos20^\circ\cos50^\circ-\sin20^\circ\sin50^\circ}{\sin20^\circ\sin50^\circ}=\frac{\cos70^\circ}{\sin20^\circ\sin50^\circ}.$$
Multiplying by $\sin50^\circ$ leaves $\dfrac{\cos70^\circ}{\sin20^\circ}$, and $\cos70^\circ=\sin20^\circ$, so the value is $1$. The complementary pair is $20^\circ$ and $70^\circ$ — spotting which pair is complementary is the whole problem.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.