Question

In a triangle ABC, $\frac{a(a+c-b)}{b(b+c-a)}$ is equal to

• A. $\frac{(1–\cos A)}{(1–\cos B)}$
• B. $\frac{(1+\cos B)}{(1+\cos A)}$
• C. $\frac{(1–\cos A)}{(1+\cos B)}$
• D. None of these

Answer 1

The correct option is A

$\frac{(1–\cos A)}{(1–\cos B)}$

Explanation for the correct option:

Finding the value of the expression:

Given,

$\frac{a(a+c-b)}{b(b+c-a)}$

Multiplying and dividing the above expression by $(a+b+c)$

$\begin{array}{rcl}\frac{a(a+c-b)}{b(b+c-a)}& =& \frac{a(a+c–b)(a+c+b)}{b(b+c–a)(b+c+a)}\\ & =& \frac{a({(a+c)}^2–b^2)}{b({(b+c)}^2–a^2)}\\ & =& \frac{a(a^2+c^2+2ac–b^2)}{b(b^2+c^2+2bc–a^2)}\\ & =& \frac{a(2acC\mathrm{os}B+2ac)}{b(2bcC\mathrm{os}A+2bc)}\left[\mathbf{by}\mathbf{}\mathbf{cosine}\mathbf{}\mathbf{rule}\right]\\ & =& \frac{2a^{2}c(C\mathrm{os}B+1)}{2b^{2}c(C\mathrm{os}A+1)}\\ & =& \frac{{\sin }^{2}A(1+\cos B)}{{\sin }^{2}B(1+\cos A)}\left[\mathbf{by}\mathbf{}\mathbf{sine}\mathbf{}\mathbf{rule}\right]\\ & =& \frac{(1–{\cos }^{2}A)(1+\cos B)}{(1–{\cos }^{2}B)(1+\cos A)}\\ & =& \frac{(1–\cos A)}{(1–\cos B)}\mathbf{[}\mathbf{\because }\mathbf{}\left(a^2-b^2\right)\mathbf{=}\left(a+b\right)\left(a-b\right)\mathbf{]}\end{array}$

Hence, the correct option is A.