InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
यदि `ax + b {sec(tan^(-1)x)} = c` तथा `ay + b {sec(tan^(-1)y)} = c`, तब ` (x+y)/(1-xy)` का मान होगाA. `(2ab)/(1^(2)-c^(2))`B. `(2ac)/(a^(2)-c^(2))`C. `(c^(2)-b^(2))/(a^(2)+b^(2))`D. इनमें से कोई नहीं |
|
Answer» Correct Answer - B 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)` |
|