Question

If tan θ = `1/sqrt(7) `show that  ` (cosec ^2  θ - sec^2 θ)/(cosec^2 θ + sec^2 θ ) = 3/4` 

Answer 1

Let us consider a right ΔABC, right angled at B and ∠𝐶 = 𝜃.
Now it is given that tan 𝜃 = `(AB)/(BC) = 1/(sqrt(7))`

So, if AB = k, then BC = `sqrt(7) `𝑘, 𝑤ℎ𝑒𝑟 𝑘 𝑖𝑠 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟.
Using Pythagoras theorem, we have:
`AC^2 = AB^2 + BC^2`
`⟹ AC^2 = (k)^2 + (sqrt(7K)`
`⟹ AC^2 = K^2 + 7K^2`
`⟹ AC = 2sqrt(2K)`
Now, finding out the values of the other trigonometric ratios, we have:

sin θ = `(AB)/(AC) = K/(2sqrt(2k)) = 1/(2sqrt(2))`

cos  θ  = `( BC)/(AC) = (sqrt(7k))/(2 sqrt(2k)) = (sqrt(7))/(2sqrt(2))`

∴ 𝑐𝑜𝑠𝑒𝑐 𝜃 =`1/(sin θ) = 2 sqrt(2) and sec θ = 1/cos θ = (2sqrt(2))/(sqrt(7))`

Substituting the values of cosec θ and sec θ in the give expression, we get:

`(cosec^2  θ - sec^2 θ)/(cosec^2 θ + sec2 θ)`

=`(2 sqrt(2)^2-((2sqrt(2))/sqrt(7))^2)/(2sqrt({2)^2)+((2sqrt(2))/sqrt(7))^2)`

=`(8-(8/7))/(8+(8/7))`

= `((56-8)/7)/((56+8)/7)`

=`48/64 = 3/4 `= 𝑅𝐻𝑆
i.e., LHS = RHS
Hence proved.