Without using trigonometric tables, evaluate :

$\Big(\dfrac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 10° - \text{tan}^2 80°}\Big).$

Solving,

$\Rightarrow \Big(\dfrac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 \space 10° - \text{tan}^2 \space 80°}\Big) \\[1em] \Rightarrow \Big(\dfrac{\text{sin 35° cos (90° - 35°) + cos 35° sin (90° - 35°)}}{\text{cosec}^2 \space (90° - 80°) - \text{tan}^2 \space 80°}\Big) \\[1em] \text{As, sin (90 - θ) = cos θ, cos (90 - θ) = sin θ and cosec (90 - θ) = sec θ} \\[1em] \Rightarrow \Big(\dfrac{\text{sin 35° sin 35° + cos 35° cos 35°}}{\text{sec}^2 \space 80° - \text{tan}^2 \space 80°}\Big) \\[1em] \Rightarrow \Big(\dfrac{\text{sin}^2 \space 35° + \text{cos}^2 \space 35°}{\text{sec}^2 \space 80° - \text{tan}^2 \space 80°}\Big) \\[1em] \text{As, sin}^2 θ + \text{cos}^2 θ = 1 \text{ and } \text{sec}^2 θ - \text{tan}^2 θ = 1 \\[1em] \Rightarrow \dfrac{1}{1} \\[1em] \Rightarrow 1.$

Hence, $\Big(\dfrac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 \space 10° - \text{tan}^2 \space 80°}\Big)$ = 1.