Question

If in a $∆\mathrm{ABC}$, $\frac{a}{\cos A}=\frac{b}{\cos B}$, then

• A. ${\sin }^{2}A+{\sin }^{2}B=\sin 2C$
• B. $2\sin A\cos B=\sin C$
• C. $2\sin A\sin B\sin C=1$
• D. None of the above.

Answer 1

The correct option is B

$2\sin A\cos B=\sin C$

Explanation for the correct option:

Find the relation:

Given,

$\frac{a}{\cos A}=\frac{b}{\cos B}$

Concept: We know the cosine rule,

$$\begin{gathered} a^2=b^2+c^2–2bc\cos A, \\ {b}^2=a^2+c^2–2ac\cos B, \\ {c}^2=a^2+b^2–2ab\cos C \end{gathered}$$

$\begin{array}{rcl}\frac{a}{\cos A}& =& \frac{b}{\cos B}\\ & \Rightarrow & a\cos A=b\cos B\\ & \Rightarrow & a\frac{(a^2+c^2-b^2)}{2ac}=b\frac{(b^2+c^2-a^2)}{2bc}\\ & \Rightarrow & a^2+c^2-b^2=b^2+c^2-a^2\\ & \Rightarrow & 2a^2=2b^2\\ & \Rightarrow & a^2=b^2\end{array}$

Now, find the value of $2\sin A\cos B$.

$\begin{array}{rcl}2\sin A\cos B& =& 2\times \frac{a}{k}\times \frac{(a^2+c^2-b^2)}{2ac}\mathbf{[}\mathbf{\because }\frac{\mathbf{a}}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{A}}\mathbf{=}\frac{\mathbf{b}}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{B}}\mathbf{=}\frac{\mathbf{c}}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{C}}\mathbf{=}\mathbf{k}\mathbf{]}\\ & =& 2\times \frac{a}{k}\times \frac{(b^2+c^2-b^2)}{2ac}\mathbf{[}{\mathbf{a}}^{\mathbf{2}}\mathbf{=}{\mathbf{b}}^{\mathbf{2}}\mathbf{,}\mathbf{}\mathbf{a}\mathbf{b}\mathbf{o}\mathbf{v}\mathbf{e}\mathbf{}\mathbf{p}\mathbf{r}\mathbf{o}\mathbf{v}\mathbf{e}\mathbf{d}\mathbf{]}\\ & =& \frac{c^2}{ck}\\ & =& \frac{c}{k}\\ & =& \sin C\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\frac{\mathbf{a}}{\mathbf{sinA}}\mathbf{=}\frac{\mathbf{b}}{\mathbf{sinB}}\mathbf{=}\frac{\mathbf{c}}{\mathbf{sinC}}\mathbf{=}\mathbf{k}\mathbf{]}\end{array}$

Hence, the correct option is B.