Question

`sqrt(-4 + sqrt(8 + 16  "cosec"^4 θ + sin^4 θ)) = A  "cosec"  θ + B sin θ`, then A = ______ and B = ______.

Answer 1

`sqrt(-4 + sqrt(8 + 16  "cosec"^4 θ + sin^4 θ)) = A  "cosec"  θ + B sin θ`, then A = 2 and B = –1.

Explanation:

Let x = sin θ (x ≠ 0).

Then `sqrt(8 + 16  "cosec"^4 θ + sin^4 θ)`

= `sqrt(x^4 + 16/x^4 + 8)` 

= `x^2 + 4/x^2`

Since `(x^2 + 4/x^2)^2 = x^4 + 16/x^4 + 8`. 

Hence the given expression becomes `sqrt(-4 + x^2 + 4/x^2) = sqrt((x - 2/x)^2) = |x - 2/x|`. 

For 0 < θ < π (so sin θ > 0) this equals `2/x - x = 2  "cosec"  θ - sin θ`, so A = 2 and B = –1. (If sin θ < 0 the sign reverses, giving A = –2, B = 1.)