UPSEE Solved Model Paper 6

We can solve this problem using the basic formula of trigonometric function. Basic formula $\sin^2 A + \cos^2 A$ = 1 $\cos^2 A$ = 1 - $\sin^2 A$    By using this formula we can find the value of k. We have $\frac{\cos A + \cos B}{\sin A + \sin B}$ = $k\left(\frac{\sin A - \sin B}{\cos A - \cos B}\right)$ $\left(\text{Since } \frac{a}{b} = k \frac{c}{d}\right)$ ⇒ $\frac{\cos A + \cos B}{\sin A + \sin B}$ × $\frac{\cos A - \cos B}{\sin A - \sin B}$ = k [Since If $\frac{a}{b}$ = $k \frac{c}{d}$ , $\frac{ad}{bc}$ = k) ⇒ $\frac{\cos^2 A - \cos^2 B}{\sin^2 A - \sin^2 B}$ = k (Since (a + b) (a - b) = $a^2 - b^2$) Cross multiplying we get $\cos^2 A - \cos^2 B$ = $k(\sin^2 A - \sin^2 B)$ $(1 - \sin^2 A) - (1 - \sin^2 B)$ = $k(\sin^2 A - \sin^2 B)$ $1 - \sin^2 A - 1 + \sin^2 B$ = $k(\sin^2 A - \sin^2 B)$ $-\sin^2 A + \sin^2 B$ = $k(\sin^2 A - \sin^2 B)$ $-1(\sin^2 A - \sin^2 B)$ = $k(\sin^2 A - \sin^2 B)$ i.e., $-1(\sin^2 A - \sin^2 B)$ = $\frac{\cos A + \cos B}{\sin A + \sin B}$ ∴ k = - 1 Hence, If $k\left(\frac{\sin A - \sin B}{\cos A - \cos B}\right)$ = $k({sinA - sinB}/{cosA - cosB})$ then k is - 1