Question

The value of $e^{{\log }_{10}\tan 1+{\log }_{10}\tan 2+{\log }_{10}\tan 3+...+{\log }_{10}\tan 89}$ is

• A. $0$
• B. $1$
• C. $\frac{1}{e}$
• D. $e$

Answer 1

Answer: (B)

$1$

We know, $\tan \theta \tan (90-\theta )=\tan \theta \cot \theta$

$=1$

ie $\tan 1^o\tan {89}^o=\tan 2^o\tan {88}^o=\tan 3^o\tan {87}^o$

$........=\tan {44}^o\tan {46}^o=1$

So, ${\log }_{10}\tan 1^o+{\log }_{10}\tan 2^o....{\log }_{10}tan{89}^o$

$=({\log }_{10}\tan 1^o+{\log }_{10}\tan {89}^o)+({\log }_{10}\tan 2^o+{\log }_{10}\tan {88}^o)$

$+....+({\log }_{10}\tan {44}^o+{\log }_{10}\tan {46}^o)+{\log }_{10}\tan {45}^o$

$={\log }_{10}\left(\tan 1^o\tan {89}^o\right)+{\log }_{10}\left(\tan 2^o\tan {88}^o\right)$

$+.....+{\log }_{10}\left(\tan {44}^o\tan {46}^o\right)+{\log }_{10}tan{45}^o$

$={\log }_{10}1+{\log }_{10}1+......{\log }_{10}1+{\log }_{10}1$

$=0$, as ${\log }_{10}1=0$

$\therefore$ Answer $=e^0=1$