​If $$a sec A + b tan A + c = 0$$​ and $$a' sec A + b' tan A + c' = 0,$$​ then $$(bc'—bc)^2-(ca'-c'a)^2$$ is equal to:​

Correct option is D

​Given:

​$$a \sec A + b \tan A + c = 0 ......(i) \\ \ \\ a' \sec A + b' \tan A + c' = 0........(ii)$$​​

to evaluate:
​$$(bc' - b'c)^2 - (ca' - c'a)^2$$​​

Formula Used: 

$$\sec^2 A - \tan^2 A = 1$$​

​$$(X)^2 - (Y)^2 = (X - Y)(X + Y)$$​​

Solution:

Multiply (i) by (a') and (ii) by (a):

a' a sec A + a' b tan A + a' c = 0...........{iii}

a a' sec A + a b' tan A + a c' = 0 ..........{iv}

Subtract (iv) – (iii):

a b' tan A - a' b tan A + a c' - a' c = 0

(a b' - a' b) tan A + (a c' - a' c) = 0

-(a b' - a' b)tan A = (a' c - a c') ​..........{v}

Similarly, eliminate (tan A) to get (sec A):

Multiply (i) by (b'), (ii) by (b):

a b' sec A + b b' tan A + b' c = 0

a' b sec A + b' b tan A + b c' = 0

Subtracting above equation:

(a b' - a' b) sec A + (b' c - b c') = 0

-(a b' - a' b)sec A = (b c' - b' c).......{vi}

Now

​$$(b c' - b' c)^2 - (c a' - c' a)^2$$​​

Let X = b c' - b' c,  Y = c a' - c' a

Then:

​$$X^2 - Y^2 = (b c' - b' c)^2 - (c a' - c' a)^2$$​​

Hence, from equation (v) and (vi)

​$$(b c' - b' c)^2 - (c a' - c' a)^2 = (a b' - a' b)^2$$ ​