A) \[{{a}^{2}}-{{b}^{2}}\]
B) \[{{b}^{2}}-{{a}^{2}}\]
C) \[{{a}^{2}}+{{b}^{2}}\]
D) \[b-a\]
Correct Answer: B
Solution :
| [b] We have, \[a\cot \theta +b\cos ec\,\theta =p\] ...(1) |
| \[b\cot \theta +a\cos ec\,\theta =q\] ...(2) |
| Squaring and then subtracting eq. (2) from eq. (1), we get |
| \[\therefore \,{{p}^{2}}-{{q}^{2}}={{a}^{2}}{{\cot }^{2}}\theta +{{b}^{2}}\cos e{{c}^{2}}\theta +2ab\,\cot \theta \,\cos ec\,\theta \] |
| \[-{{b}^{2}}{{\cot }^{2}}\theta -{{a}^{2}}\cos e{{c}^{2}}\theta -2ab\cot \theta \,\,\text{cosec}\theta \] |
| \[={{a}^{2}}({{\cot }^{2}}\theta -\cos e{{c}^{2}}\theta )+{{b}^{2}}(\cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta )\] |
| \[=-{{a}^{2}}+{{b}^{2}}\]\[[\cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta =1]\] |
| \[={{b}^{2}}-{{a}^{2}}\] |
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