Question

Show that $\frac{co{s}^2({45}^{\circ }+\theta )+co{s}^2({45}^{\circ }-\theta )}{tan({60}^{\circ }+\theta )tan({30}^{\circ }-\theta )}=1$

Answer 1

$LHS=\frac{co{s}^2({45}^{\circ }+\theta )+co{s}^2({45}^{\circ }-\theta )}{tan({60}^{\circ }+\theta ).tan({30}^{\circ }-\theta )}$

$=\frac{co{s}^2({45}^{\circ }+\theta )+{\left[sin\right\{{90}^{\circ }-({45}^{\circ }-\theta )\left\}\right]}^2}{tan({60}^{\circ }+\theta ).cot\{{90}^{\circ }-({30}^{\circ }-\theta \left)\right\}}$

$[\because sin({90}^{\circ }-\theta )=cos\theta andcot({90}^{\circ }-\theta )=tan\theta ]$

$=\frac{co{s}^2({45}^{\circ }+\theta )+si{n}^2({45}^{\circ }+\theta )}{tan({60}^{\circ }+\theta ).cot({60}^{\circ }+\theta )}[\because si{n}^2\theta +co{s}^2\theta ]$

$=\frac{1}{tan({60}^{\circ }+\theta )\frac{1}{tan({60}^{\circ }+\theta )}}=1=RHS$