InterviewSolution
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InterviewSolution
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यदि `"cos"^(-1)(x)/(a)+"cos"^(-1)(y)/(b)=alpha` तब सिद्ध कीजिए - `(x^(2))/(a^(2))-(2xy)/(ab)cos alpha + (y^(2))/(b^(2))=sin^(2)alpha`. |
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Answer» यहाँ `"cos"^(-1)(x)/(a)+"cos"^(-1)(y)/(b)=alpha` `rArr cos^(-1)[(x)/(a).(y)/(b)-sqrt(1-(x^(2))/(a^(2)))sqrt(1-(y^(2))/(b^(2)))]=alpha` `rArr cos^(-1)[(xy)/(ab)-sqrt(1-(x^(2))/(a^(2)))sqrt(1-(y^(2))/(b^(2)))]=alpha` `rArr " " (xy)/(ab)-sqrt(1-(x^(2))/(a^(2)))sqrt(1-(y^(2))/(b^(2)))=cos alpha` `rArr ((xy)/(ab)-cos alpha)= sqrt(1-(x^(2))/(a^(2)))sqrt(1-(y^(2))/(b^(2)))` दोनों पक्षों का वर्ग करने पर, `((xy)/(ab)-cos alpha)^(2)= (1-(x^(2))/(a^(2)))(1-(y^(2))/(b^(2)))` `rArr (x^(2)y^(2))/(a^(2)b^(2))-(2xy)/(ab)cos alpha + cos^(2)alpha` `= 1-(x^(2))/(a^(2))-(y^(2))/(b^(2))+(x^(2)y^(2))/(a^(2)b^(2))` `rArr (x^(2))/(a^(2))+(y^(2))/(b^(2))-(2xy)/(ab)cos alpha = 1- cos^(2)alpha` `rArr (x^(2))/(a^(2))+(y^(2))/(b^(2))-(2xy)/(ab)cos alpha = sin^(2)alpha`. यही सिद्ध करना था | |
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