Question

$\cos {45}^o.\tan {(-495)}^o-\tan {585}^o.\cot \left({-495}^o\right)=0$

Answer 1

$=\cos {45}^{\circ }\tan \left({-495}^{\circ }\right)-\tan \left({585}^{\circ }\right)\cot \left({-495}^{\circ }\right)$

$=-\cos {45}^{\circ }\tan {495}^{\circ }+\tan \left({585}^{\circ }\right)\cot {495}^{\circ }$ since $\tan (-\theta )=-\tan \theta$ and $\cot (-\theta )=-\cot \theta$ which are in $4^{th}$ quadrants.

$=-\frac{1}{\sqrt{2}}\tan ({360}^{\circ }+{135}^{\circ })+\tan ({360}^{\circ }+{225}^{\circ })\cot ({360}^{\circ }+{135}^{\circ })$ where $\cos {45}^{\circ }=\frac{1}{\sqrt{2}}$

$=-\frac{1}{\sqrt{2}}\tan {135}^{\circ }+\tan {225}^{\circ }\cot {135}^{\circ }$ where ${360}^{\circ }+\theta$ is in first quadrant all are positive

$=\frac{-1}{\sqrt{2}}\tan ({180}^{\circ }-{45}^{\circ })+\tan ({180}^{\circ }+{45}^{\circ })\cot ({180}^{\circ }-{45}^{\circ })$

Since ${180}^{\circ }-\theta$ is in second quadrant $\tan ({180}^{\circ }-\theta )$ and $\cot ({180}^{\circ }-\theta )$ is negative and ${180}^{\circ }+\theta$ is in third quadrant $\tan ({180}^{\circ }+\theta )$ and $\cot ({180}^{\circ }+\theta )$ are positive

$\therefore \frac{1}{\sqrt{2}}\tan {45}^{\circ }+\tan {45}^{\circ }\times -\cot {45}^{\circ }$

$=\frac{1}{\sqrt{2}}\times 1+1\times -1$ where $\tan {45}^{\circ }=1$ and $\cot {45}^{\circ }=1$

$=\frac{1}{\sqrt{2}}-1$