Trigonometric Functions — MATHS STD 11 Science — Question

Consider:

$\text{a}(\cos\text{B}\cos\text{C}+\cos\text{A})$

$=\text{k}\sin\text{A}(\cos\text{B}\cos\text{C}+\cos\text{A})$

$=\text{k}(\sin\text{A}\cos\text{B}\cos\text{C}+\cos\text{A}\sin\text{A})$

$=\text{k}\Big[\frac{1}{2}\cos\text{C}\{\sin(\text{A + B})+\sin(\text{A}-\text{B})\}+\sin\text{A}\cos\text{A}\Big]$

$=\text{k}\Big[\frac{1}{2}\{\sin(\text{A + B})\cos\text{C}+\sin(\text{A}-\text{B})\cos\text{C}\}+\sin\text{A}\cos\text{A}\Big]$

$=\text{k}\Big[\frac{1}{2}\big\{\frac{1}{2}\big[\sin(\text{A + B + C})+\sin(\text{A + B}-\text{C})\\+\sin(\text{A}-\text{B + C})+\sin(\text{A}-\text{B}-\text{C})\big]\big\}+\sin\text{A}\cos\text{A}]$

$=\text{k}\Big[\frac{1}{4}\{\sin\pi+\sin(\pi-2\text{C})+\sin(\pi-2\text{B})-\sin(\pi-2\text{A})\}+\frac{\sin2\text{A}}{2}\Big]$ $(\because\text{A + B + C} = \pi)$

$=\frac{\text{k}}{4}(\sin2\text{C}+\sin2\text{B}+\sin2\text{A})...(1)$

and

$\text{b}(\cos\text{A}\cos\text{C}+\cos\text{B})$

$=\text{k}(\sin\text{B}\cos\text{A}\cos\text{C}+\sin\text{B}\cos\text{B})$

$=\text{k}\Big[\frac{1}{2}\cos\text{A}\{\sin(\text{B + C})+\sin(\text{B} -\text{C})\}+\frac{\sin2\text{B}}{2}\Big]$

$=\text{k}\Big(\frac{1}{2}(\sin(\text{B + C})\cos\text{A}+\sin(\text{B}-\text{C})\cos\text{A})+\frac{\sin2\text{B}}{2}\Big)$

$=\text{k}\Big(\frac{1}{4}(\sin(\text{B + C + A})+\sin(\text{B + C}-\text{A})\\+\sin(\text{B}-\text{C + A})+\sin(\text{B}-\text{C}-\text{A}))+\frac{\sin2\text{B}}{2}\Big)$

$=\frac{\text{k}}{2}\Big(\sin\pi+\sin(\pi-2\text{A})+\sin(\pi-2\text{C})-\sin(\pi-2\text{B})+\frac{\sin2\text{B}}{2}\Big)$ $(\because\text{A + B + C}=\pi)$

$=\frac{\text{k}}{4}(\sin2\text{A}+\sin2\text{C}+\sin2\text{B})...(2)$

Similarly,

$\text{c}(\cos\text{A}\cos\text{B}+\cos\text{C})=\frac{\text{k}}{4}(\sin2\text{A}+\sin2\text{B}+\sin2\text{C})...(3)$

From (1), (2) and (3), we get:

$\text{a}(\cos\text{B}\cos\text{C}+\cos\text{A})=\text{b}(\cos\text{C}\cos\text{A}+\cos\text{B})\\=\text{c}(\cos\text{A}\cos\text{B}+\cos\text{C})$

Hence proved.