MCQOPTIONS
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| 1. |
If \[a{{\sin }^{2}}x+b{{\cos }^{2}}x=c,\,\,\]\[b\,{{\sin }^{2}}y+a\,{{\cos }^{2}}y=d\] and \[a\,\tan x=b\,\tan y,\]then \[\frac{{{a}^{2}}}{{{b}^{2}}}\] is equal to |
| A. | \[\frac{(b-c)\,\,(d-b)}{(a-d)\,\,(c-a)}\] |
| B. | \[\frac{(a-d)\,\,(c-a)}{(b-c)\,\,(d-b)}\] |
| C. | \[\frac{(d-a)\,\,(c-a)}{(b-c)\,\,(d-b)}\] |
| D. | \[\frac{(b-c)\,\,(b-d)}{(a-c)\,\,(a-d)}\] |
| Answer» C. \[\frac{(d-a)\,\,(c-a)}{(b-c)\,\,(d-b)}\] | |