UPSC CDS 1 2022 Math Paper

Concept:Use complementary angle identities to simplify trigonometric terms.Key identities: $\sin(90^\circ - \theta) = \cos\theta$, $\cos(90^\circ - \theta) = \sin\theta$, $\tan(90^\circ - \theta) = \cot\theta$, $\cot(90^\circ - \theta) = \tan\theta$, and $\cot\theta = \frac{1}{\tan\theta}$.Explanation:Start with the given expression:$\frac{2\sin 68^\circ}{\cos 22^\circ} - \frac{2\cot 15^\circ}{5\tan 75^\circ} - \frac{3\tan 20^\circ \tan 40^\circ \tan 45^\circ \tan 50^\circ \tan 70^\circ}{5}$Rewrite $\cos 22^\circ$ using complementary identity: $\cos 22^\circ = \cos(90^\circ - 68^\circ) = \sin 68^\circ$.Thus, first term becomes $\frac{2\sin 68^\circ}{\sin 68^\circ} = 2$.For the second term, $\tan 75^\circ = \tan(90^\circ - 15^\circ) = \cot 15^\circ$.Hence, $\frac{2\cot 15^\circ}{5\tan 75^\circ} = \frac{2\cot 15^\circ}{5\cot 15^\circ} = \frac{2}{5}$.For the third term, apply complementary pairs: $\tan 20^\circ = \tan(90^\circ - 70^\circ) = \cot 70^\circ$ and $\tan 40^\circ = \tan(90^\circ - 50^\circ) = \cot 50^\circ$.Substitute: $\tan 20^\circ \tan 40^\circ \tan 45^\circ \tan 50^\circ \tan 70^\circ = \cot 70^\circ \cot 50^\circ \cdot 1 \cdot \tan 50^\circ \cdot \tan 70^\circ$.Since $\cot 70^\circ = \frac{1}{\tan 70^\circ}$ and $\cot 50^\circ = \frac{1}{\tan 50^\circ}$, the product simplifies to $1 \cdot 1 \cdot 1 = 1$.Therefore, the third term becomes $\frac{3 \times 1}{5} = \frac{3}{5}$.Now combine all terms: $2 - \frac{2}{5} - \frac{3}{5} = 2 - \frac{5}{5} = 2 - 1 = 1$.Answer:$1$