Question

If sin θ ,` sqrt (3)/2` find the value of all T- ratios of θ .

Answer 1

Let us first draw a right ΔABC, right angled at B and ∠𝐶 = 𝜃
Now, we know that sin 𝜃 = `"𝑃𝑟𝑒𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 "/"ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒" = (AB)/ (AC) = sqrt(3)/2`

So, if AB = `sqrt(3)`𝑘, 𝑡ℎ𝑒𝑛 𝐴𝐶 = 2𝑘, 𝑤ℎ𝑒𝑟𝑒 𝑘 𝑖𝑠 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟.
Now, using Pythagoras theorem, we have:

`AC^2 = AB^2 + BC^2`
`⟹ BC^2 = AC^2 − AB^2 = (2K)^2 − (sqrt(3)K)^2`
`⟹ BC^2 = 4K^2 − 3K^2 = K^2`
⟹ BC = k
Now, finding the other T-rations using their definitions, we get:

 Cos 𝜃 =` (BC)/(AC) = K/(2K)= 1/2`

Tan 𝜃 =`(AB)/(BC) = (sqrt(3K))/K = sqrt(3)`

∴ cot 𝜃 =`1/(Tan θ ) = 1/sqrt(3), cosec   θ = 1/(sin θ) = 2/(sqrt(3)) and sec θ = 1/ (cos θ) = 2 `