The value of (sin 70 ° ) (cot 10 ° cot 70 ° -1 ) is

Mathematics · Trigonometry Ratios and Identities

JEE Main 2025 — 23 January, Morning Shift — Question 15

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

E=\left(\sin 70^\circ\right)\left(\cot 10^\circ \cot 70^\circ-1\right)

\cot 10^\circ=\frac{\cos 10^\circ}{\sin 10^\circ},\quad \cot 70^\circ=\frac{\cos 70^\circ}{\sin 70^\circ}

Using

\cos 70^\circ=\sin 20^\circ,\quad \sin 70^\circ=\cos 20^\circ

\cot 10^\circ \cot 70^\circ = \frac{\cos 10^\circ \sin 20^\circ}{\sin 10^\circ \cos 20^\circ} = \frac{\tan 20^\circ}{\tan 10^\circ}

E=\sin 70^\circ\left(\frac{\tan 20^\circ}{\tan 10^\circ}-1\right)

Since $\sin 70^\circ=\cos 20^\circ$ ,

E=\cos 20^\circ\left(\frac{\tan 20^\circ}{\tan 10^\circ}-1\right)

Using the identity

\tan 20^\circ=\frac{2\tan 10^\circ}{1-\tan^2 10^\circ}

\frac{\tan 20^\circ}{\tan 10^\circ} =\frac{2}{1-\tan^2 10^\circ}

\frac{2}{1-\tan^2 10^\circ}-1 =\frac{1+\tan^2 10^\circ}{1-\tan^2 10^\circ}

Also,

\cos 20^\circ=\frac{1-\tan^2 10^\circ}{1+\tan^2 10^\circ}

E=\frac{1-\tan^2 10^\circ}{1+\tan^2 10^\circ} \cdot \frac{1+\tan^2 10^\circ}{1-\tan^2 10^\circ} =1

\boxed{1}

Answer key and solution verified before publishing.

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