For any Δ ABC show that: (c - b cos A)/(b - c cos A) = cos B/cos C

1.	For any Δ ABC show that: (c - b cos A)/(b - c cos A) = cos B/cos C
Answer» Let us consider the LHS (c - b cos A)/(b - c cos A) Now, we can observe that we can get terms c – b cos A and b – c cos A from projection formula From the projection formula we get, c = a cos B + b cos A c – b cos A = a cos B …. (i) And, b = c cos A + a cos C b – c cos A = a cos C …. (ii) On dividing equation (i) by (ii), we get, (c - b cos A)/(b - c cos A) = a cos B/a cos C = cos B/cos C = RHS Thus proved.

Answer»

Let us consider the LHS

(c - b cos A)/(b - c cos A)

Now, we can observe that we can get terms c – b cos A and b – c cos A from projection formula

From the projection formula we get,

c = a cos B + b cos A

c – b cos A = a cos B …. (i)

And,

b = c cos A + a cos C

b – c cos A = a cos C …. (ii)

On dividing equation (i) by (ii), we get,

(c - b cos A)/(b - c cos A)

= a cos B/a cos C

= cos B/cos C

= RHS

Thus proved.

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