Question

If `cot theta = 1/sqrt3` show that  `(1 - cos^2 theta)/(2 - sin^2  theta) = 3/5`

Answer 1

`cot theta = 1/sqrt3 (1 - cos^2 theta)/(2 - sin^2 theta) = 3/5`

`cot theta = "𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒"/"𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒" = 1/sqrt3`

Let x be the hypotenuse

By applying Pythagoras

𝐴𝐶² = 𝐴𝐵² + 𝐵𝐶²

`x^2 = (sqrt3)^2 + 1`

`x^2 = 3 + 1`

𝑥² = 3 + 1 ⇒ 𝑥 = 2

`cos theta = (BC)/(AC) = 1/2`

`sin theta = (AB)/(AC) = sqrt3/2`

`(1 - cos^2 theta)/(2 - sin^2 theta) => (1 - (1/2)^2)/(2 - (sqrt3)/2)^2`

`=> (1 - 1/4)/(2 - 3/4) => (3/4)/(5/4`

`= 3/5`