JEE Main 2021 (Online) 4th August Morning ShiftQuestion 1
The value of $\frac{\sum_{k=1}^{22}(-1)^{k-1}\cos k^{\circ}+\sum_{k=23}^{44}(-1)^{k}\cos k^{\circ}}{\sum_{k=1}^{22}(-1)^{k-1}\sin k^{\circ}+\sum_{k=23}^{44}(-1)^{k}\sin k^{\circ}}$, where $k^{\circ}$ denotes $k$ degree, is
Trigonometric Ratio & Identities
Solution
Concept. Pair the term $k$ with the term $45-k$: they have the same sign, and sum-to-product gives a common factor. Formulas used. (1) $\cos u+\cos v=2\cos\frac{u+v}{2}\cos\frac{u-v}{2}$, $\sin u+\sin v=2\sin\frac{u+v}{2}\cos\frac{u-v}{2}$(2) $\cot22.5^\circ=\sqrt2+1$Step 1 - Signs. For $k\in\{1,\dots,22\}$ the term $45-k\in\{23,\dots,44\}$ has sign $(-1)^{45-k}=(-1)^{k-1}$, the same as term $k$. Step 2 - Pair. By (1), with $u=k^\circ$, $v=(45-k)^\circ$: $$\text{numerator}=2\cos22.5^\circ\,T,\qquad\text{denominator}=2\sin22.5^\circ\,T,$$ where $T=\sum_{k=1}^{22}(-1)^{k-1}\cos(k-22.5)^\circ\ne0$. Step 3 - Value. The ratio is $\cot22.5^\circ=\sqrt2+1$. Answer: (D)