InterviewSolution
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InterviewSolution
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यदि `cosy=xcos(a+y),` जहाँ `cosanepm1`, तब सिद्ध कीजिए कि- `(dy)/(dx)=(cos^(2)(a+y))/(sina)` |
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Answer» यहाँ `cosy=xcos(a+y)` `rArrx=(cosy)/(cos(a+y))` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, `(dy)/(dx)=(cos(a+y)d/(dy)(cosy)-cosyd/(dy)cos(a+y))/(cos^(2)(a+y))` `rArr(dy)/(dx)=(cos(a+y)(-siny)-(cosy)(-sin(a+y))d/(dy)(a+y))/(cos^(2)(a+y))` `rArr(dy)/(dx)=(-cos(a+y)siny+sin(a+y)cosy)/(cos^(2)(a+y))` `rArr(dy)/(dx)=(sin(a+y)cosy-cos(a+y)siny)/(cos^(2)(a+y))` `(dy)/(dx)=(sin(a+y-y))/(cos^(2)(a+y))` `[becausesinAcosB-cosAsinB=sin(A-B)]` `rArr(dy)/(dx)=(sina)/(cos^(2)(a+y))` `rArr(dy)/(dx)=(cos^(2)(a+y))/(sina)` यही सिद्ध करना था |
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