यदि `cos theta = 3/5,`तब `(sin theta - 1/(tan theta))/(2 tan theta)`का मान परिकलित कीजिए ।

Explanation

हम एक समकोण `Delta ABC` खींचते है , जिसमें `angle B = 90^(@) , BC = 3, AC = 5`
अब , `cos angleACB = (BC)/(AC) = 3/5`
, दिया है ,`cos theta = 3/5`
` :. cos angleACB = cos theta`
` :. angleACB = theta`
समकोण `Delta ABC` में, `p = sqrt(h^(2)-b^(2)) = sqrt((5)^(2)-(3)^(2)) = sqrt(25 -9) = sqrt(16) = 4`
`:. sin theta = p/h = 4/5 : tan theta = p/h = 4/3 :. 1/(tan theta) = 3/4`
अब , `(sin theta - 1/(tan theta))/(2 tan theta) = (4/5 - 3/4)/(2 xx 4/3) = ((16-15)/20)/(8/3) = (1/20)/(8/3) = 1/20 xx 3/8 = 3/160`