यदि `ax + b {sec(tan^(-1)x)} = c` तथा `ay + b {sec(tan^(-1)y)} = c`, तब
xy का मान होगा

Options

A. `(2ab)/(a^(2)-b^(2))`
B. `(c^(2)-b^(2))/(a^(2) -b^(2))`
C. `(c^(2)-b^(2))/(a^(2)+b^(2))`
D. इनमें से कोई नहीं

Answer

B

Explanation

मान लीजिए `tan^(-1) x = alpha` तथा `tan^(-1) y = beta`
तब , `a tan alpha + b sec alpha = c ` ...(i)
तथा `a tan beta+ b sec beta = c ` ...(ii)
समी (i) और (ii) से स्पष्ट है कि समीकरण `a tan theta + b sec theta = c ` के मूल `alpha`तथा `beta`है ।
अब ,`a tan theta + b sec theta = c `
` rArr b sec theta = c - a tan theta`
` rArr b^(2) sec^(2) theta = c^(2) - 2ac tan theta + a^(2) tan^(2) theta`
` rArr b^(2) + b^(2) tan^(2) theta = c^(2) - 2ac tan theta + a^(2) tan ^(2) theta`
` rArr (a^(2) - b^(2)) tan^(2) theta - 2ac tan theta + c^(2) - b^(2) = 0`
` rArr tan alpha+ tan beta = x + y = (2ac)/(a^(2) - b^(2))`
तथा `tan alpha* tan beta = xy = (c^(2) - b^(2))/(a^(2) - b^(2))`
` (x+y)/(1-xy)=((2ac)/(a^(2) -b^(2)))/(1-(c^(2)-b^(2))/(a^(2) -b^(2)))=(2ac)/(a^(2) -c^(2))` तथा `(1 + xy)/(x+y) = (1+(c^(2)-b^(2))/(a^(2)-b^(2)))/((2ac)/(a^(2)-b^(2)))`
` = (a^(2) + c^(2) - 2b^(2))/(2ac)`