Trigonometry – Compound angle formula – 23 January 2025 (Shift 1)

Question ID: #244
JEE Main23 January Shift 1, 2025Trigonometry

The value of $(\sin 70^{\circ})(\cot 10^{\circ}\cot 70^{\circ}-1)$ is

  • (1) 1
  • (2) 0
  • (3) $3/2$
  • (4) $2/3$

Solution:


Expression: $E = \sin 70^{\circ}\left(\frac{\cos 10^{\circ}}{\sin 10^{\circ}} \cdot \frac{\cos 70^{\circ}}{\sin 70^{\circ}} – 1\right)$

Take LCM inside the bracket:
$E = \sin 70^{\circ} \left( \frac{\cos 10^{\circ}\cos 70^{\circ} – \sin 10^{\circ}\sin 70^{\circ}}{\sin 10^{\circ}\sin 70^{\circ}} \right)$

Simplify numerator using $\cos(A+B)$:
Numerator $= \cos(10^{\circ} + 70^{\circ}) = \cos 80^{\circ}$.

$E = \sin 70^{\circ} \cdot \frac{\cos 80^{\circ}}{\sin 10^{\circ}\sin 70^{\circ}}$

$E = \frac{\cos 80^{\circ}}{\sin 10^{\circ}}$

Since $\cos 80^{\circ} = \cos(90^{\circ}-10^{\circ}) = \sin 10^{\circ}$:
$E = \frac{\sin 10^{\circ}}{\sin 10^{\circ}} = 1$.

Ans. (1)

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