Question

If $h\left(x\right)={\log }_{10}x$ then the value of $\sum _{n=1}^{89}h\left(\tan n^o\right)=$

• A. $1$
• B. $0$
• C. $-1$
• D. None of the above

Answer 1

The correct option is B $0$

$\sum _{89}^{n=1}h\left(\tan \right(n^{\circ }\left)\right)$

$=\log \left(\tan 1^{\circ }\right)+\log \left(\tan 2^{\circ }\right)+......\log \left({89}^{\circ }\right)$

$=\log \left(\right(\tan 1^{\circ }\left)\right(\tan 2^{\circ }\left)\right(\tan 3^{\circ }).........(\tan {89}^{\circ }\left)\right)$

using $\tan (90-\theta )=\cot \theta$ and $\tan \theta .\cot \theta =1$

$\therefore =\log \left(\right(\tan 1^{\circ }).....(\tan {45}^{\circ }\left)\right(\cot {44}^{\circ })-(\cot 1^{\circ }\left)\right)$

$=\log \left(1\right)$

$=0$