Evaluate :

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

Solving,

$\Rightarrow \dfrac{\text{sin 35° cos 55° + cos 35° sin 55°}}{\text{cosec}^2 10° - \text{tan}^2 80°} \\[1em] \Rightarrow \dfrac{\text{sin 35° cos (90° - 35°) + cos 35° sin (90° - 35°)}}{\text{cosec}^2 10° - \text{tan}^2 (90° - 10°)}$

By formula,

sin(90° - θ) = cos θ, cos(90° - θ) = sin θ and tan(90° - θ) = cot θ.

$\Rightarrow \dfrac{\text{sin 35° sin 35° + cos 35° cos 35°}}{\text{cosec}^2 10° - \text{cot}^2 10°} \\[1em] \Rightarrow \dfrac{\text{sin}^2 35° + \text{cos}^2 35°}{\text{cosec}^2 10° - \text{cot}^2 10°}$

By formula,

sin² θ + cos² θ = 1 and cosec² θ - cot² θ = 1.

$\Rightarrow \dfrac{1}{1}$

⇒ 1.

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