Question

Solve: $se{c}^{2}45°+co{t}^{2}135°$

Answer 1

Simplification of expresssion:

$se{c}^{2}45+co{t}^{2}135$,

$=se{c}^{2}45°+co{t}^2\left(180°-45°\right)$

$=se{c}^{2}45°+{\left(-cot45°\right)}^2\mathbf{\left[}\mathbf{\because }\mathbf{cot}\mathbf{(}\mathbf{180}\mathbf{-}\mathbf{\alpha }\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{cot\alpha }\mathbf{\right]}$

$=se{c}^{2}45°+co{t}^{2}45°$

$={\left(\sqrt{2}\right)}^2+{\left(1\right)}^2$ $\left[\because sec45°=\sqrt{2},cot45°=1\right]$

$=2+1$

$=3$

Hence, the value of $\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\mathbf{45}\mathbf{°}\mathbf{+}\mathit{c}\mathit{o}{\mathit{t}}^{\mathbf{2}}\mathbf{135}\mathbf{°}$ is $\mathbf{3}$