Question

If $X=\tan 1^0+\tan 2^0+........+\tan {45}^0$ and $y=-(\cot {46}^0+\cot {47}^0+.......+\cot {89}^0)$ then find the value of $(x+y)$.

• A. $1$
• B. $0$
• C. $-1$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

Answer: (A)

$1$

Since $\tan ({90}^o-x)=\cot {90}^0$ we have
$x+y=\tan 1^o+\tan 2^o+.......+\tan {45}^0-(\cot {46}^0+\cot {47}^0+...+\cot {89}^0)$
$=>x+y=\tan 1^o+\tan 2^o+....+\tan {45}^0-\left(\cot \right({90}^0-{44}^0)+\cot ({90}^0-{43}^0)+..+\cot ({90}^0-1^0)$
$=>x+y=\tan 1^o+\tan 2^o+....+\tan {45}^0-(\tan {44}^0+\tan {43}^0+..+\tan 1^0)$
$=>x+y=0+0+..+\tan {45}^0=1$