Without using trigonometric tables, evaluate the following:

$\dfrac{\text{sec 17°}}{\text{cosec 73°}} + \dfrac{\text{tan 68°}}{\text{cot 22°}} + \text{cos}^2\space 44° + \text{cos}^2\space 46°.$

We need to find the value of

$\dfrac{\text{sec 17°}}{\text{cosec 73°}} + \dfrac{\text{tan 68°}}{\text{cot 22°}} + \text{cos}^2\space 44° + \text{cos}^2\space 46°$

The above equation can be written as,

$\dfrac{\text{sec 17°}}{\text{cosec (90 - 17)°}} + \dfrac{\text{tan 68°}}{\text{cot (90 - 68)°}} + \text{cos}^2\space (90 - 46)° + \text{cos}^2\space 46°$

As, cos(90 - θ) = sin θ, cosec(90 - θ) = sec θ, cot(90 - θ) = tan θ and sin²θ + cos²θ = 1. Using in above equation we get,

$\Rightarrow \dfrac{\text{sec 17°}}{\text{sec 17°}} + \dfrac{\text{tan 68°}}{\text{tan 68°}} + \text{sin}^2\space46° + \text{cos}^2\space46° \\[1em] = 1 + 1 + 1 \\[1em] = 3$

Hence, the value of the expression is 3.