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рдкреНрд░рд╢реНрди
if `tan theta = 12/13` Find `(2 sin theta cos theta)/(cos^2 theta - sin^2 theta)`
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рдЙрддреНрддрд░
Let x be, the hypotenuse
By Pythagoras we get
ЁЭР┤ЁЭР╢2 = ЁЭР┤ЁЭР╡2 + ЁЭР╡ЁЭР╢2
ЁЭСе2 = 144 + 169
`x = sqrt313`
`sin theta = (AB)/(AC) = 12/sqrt313`
`cos theta = (BC)/(AC) = 13/sqrt313`
Substitute, Sin ЁЭЬГ, cos ЁЭЬГ in equation we get
`(2 sin theta cos theta)/(cos^2 theta - sin^2 theta) => (2 xx 12/sqrt313 xx 13/sqrt313)/(169/313 - 144/313)`
`= (312/313)/(25/313) = 312/25`
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