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Question Number 102178 by dw last updated on 07/Jul/20

	Answered by 1549442205 last updated on 07/Jul/20

	$Putting x = \tan α, y = \tan β we have$ $\sqrt{1 + x^{2}} = \sqrt{1 + \tan^{2} α} = \frac{1}{\cos α}, \sqrt{1 + y^{2}} = \frac{1}{\cos β}$ $x \sqrt{1 + y^{2}} + y \sqrt{1 + x^{2}} = \frac{tam α}{\cos β} + \frac{\tan β}{\cos α} = \frac{\sin α + \sin β}{\cos α \cos β} . Hence,$ $LHS = \frac{1}{\cos α \cos β} \sqrt{1 + {(\frac{\sin α + \sin β}{\cos α \cos β})}^{2}} + \frac{1}{\cos α \cos β} \sqrt{1 + {(\frac{\sin α - \sin β}{\cos α \cos β})}^{2}}$ $= \frac{1}{\cos α \cos β} \sqrt{\cos^{2} α \cos^{2} β + {(\sin α + \sin β)}^{2}} + \frac{1}{\cos α \cos β} \sqrt{\cos^{2} α \cos^{2} β + {(\sin α - \sin β)}^{2}}$ $we have : \cos^{2} α \cos^{2} β = (1 - \sin^{2} α) (1 - \sin^{2} β)$ $= 1 + \sin^{2} α \sin^{2} β - \sin^{2} α - \sin^{2} β, so$ $= \sqrt{\cos^{2} α \cos^{2} β + {(\sin α + \sin β)}^{2}} = \sqrt{1 + \sin^{2} α \sin^{2} β + 2 \sin α \sin β}$ $= \sqrt{{(\sin α \sin β + 1)}^{2}} = 1 + \sin α \sin β$ $Similarly, = \sqrt{\cos^{2} α \cos^{2} β + {(\sin α - \sin β)}^{2}} = 1 - \sin α \sin β$ $Therefore, LHS = \frac{1}{\cos α \cos β} [(1 + \sin α \sin β) - (1 - \sin α \sin β)]$ $= \frac{2 \sin α \sin β}{\cos α \cos β} = 2 \tan α \tan β = 2 xy$ $Thus, LHS = RHS (q . e . d)$
	Commented by dw last updated on 07/Jul/20

	$T h a n k y o u! Y o u r s o l u t i o n i s v e r y g o o d .$
	Commented by dw last updated on 16/Jul/20

	$b u t I d i d n o t u n d e r s t a n d!$ $\frac{1}{c o s α c o s β} [(1 + s i n α s i n β) - (1 - s i n α s i n β)] =$ $\frac{1}{c o s α c o s β} [(1 + s i n α s i n β) + (1 - s i n α s i n β)$
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