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Question
if `cos theta = 5/13` find the value of `(sin^2 theta - cos^2 theta)/(2 sin theta cos theta) = 3/5`
Solution
We have
`cos theta == 5/13`
In Δ ABC
`AC^2 = AB^2 + BC^2`
`=>(13)^2 = (AB)^2 + (5)^2`
`=> 169 = (AB)^2 + 25`
`=> (AB)^2 = 169 - 25`
=> AB = 12
`:. sin theta = 12/13 and tan theta = 12/5`
Now
`(sin^2 theta - cos^2 theta) xx 1/tan^2 theta = ((12/13)^2 - (5/13)^2)/(2 xx 12/13 xx 5/13) xx 1/(12/5)^2`
`= ((144 - 25)/169)/(120/169) xx 25/144`
`= 119/120 xx 25/144`
`= (119 xx 5)/(24 xx 144) = 595/3456`
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