Question

$\tan {27}^0\tan {32}^0+\tan {32}^0\tan {31}^0+\tan {31}^0\tan {27}^0=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

$\tan {27}^{\circ }\tan {32}^{\circ }+\tan {31}^{\circ }\tan {27}^{\circ }+\tan {32}^{\circ }\tan {31}^{\circ }$

$=\tan {27}^{\circ }(\tan {32}^{\circ }+\tan {31}^{\circ })+\tan {32}^{\circ }\tan {31}^{\circ }$

We know that $\tan A+\tan B=\tan (A+B)(1-\tan A\tan B)$

$=\tan {27}^{\circ }\left(\tan ({32}^{\circ }+{31}^{\circ })(1-\tan {32}^{\circ }\tan {31}^{\circ })\right)+\tan {32}^{\circ }\tan {31}^{\circ }$

$=\tan {27}^{\circ }\left(\tan {63}^{\circ }(1-\tan {32}^{\circ }\tan {31}^{\circ })\right)+\tan {32}^{\circ }\tan {31}^{\circ }$

$=\tan {27}^{\circ }\left(\tan ({90}^{\circ }-{27}^{\circ })(1-\tan {32}^{\circ }\tan {31}^{\circ })\right)+\tan {32}^{\circ }\tan {31}^{\circ }$

$=\tan {27}^{\circ }\cot {27}^{\circ }(1-\tan {32}^{\circ }\tan {31}^{\circ })+\tan {32}^{\circ }\tan {31}^{\circ }$

$=1-\tan {32}^{\circ }\tan {31}^{\circ }+\tan {32}^{\circ }\tan {31}^{\circ }$

$=1$