| If \[a\sec \theta =x\]and \[b\tan \theta =y,\]then how area: and y connected with a and b? [SSC (CGL) 2014] |
A) \[{{a}^{2}}{{x}^{2}}-{{b}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}\]
B) \[{{b}^{2}}{{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}\]
C) \[{{a}^{2}}{{x}^{2}}+{{b}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}\]
D) \[{{b}^{2}}{{x}^{2}}+{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}\]
Correct Answer: B
Solution :
| Given, \[a\sec \theta =x\]and \[b\tan \theta =y\] |
| By Hit and Trial, |
| \[{{b}^{2}}{{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}\] ... (i) |
| Now, putting the values of x and y in Eq. (i), we get |
| \[{{b}^{2}}{{(a\sec \theta )}^{2}}-{{a}^{2}}\,\,{{(b\tan \theta )}^{2}}={{a}^{2}}{{b}^{2}}\] |
| \[\Rightarrow \] \[{{b}^{2}}{{a}^{2}}{{\sec }^{2}}\theta -{{a}^{2}}{{b}^{2}}{{\tan }^{2}}\theta ={{a}^{2}}{{b}^{2}}\] |
| \[\Rightarrow \] \[{{a}^{2}}{{b}^{2}}({{\sec }^{2}}\theta -{{\tan }^{2}}\theta )={{a}^{2}}{{b}^{2}}\] |
| \[\therefore \] \[{{a}^{2}}{{b}^{2}}={{a}^{2}}{{b}^{2}}\] |
| \[[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1]\] |
You need to login to perform this action.
You will be redirected in
3 sec
