Question

The product $\cot {123}^o\cot {133}^o\cot {137}^o\cot {147}^o$, when simplified is equal to

• A. $-1$
• B. $\tan {37}^o$
• C. $\cot {33}^o$
• D. $1$

Answer 1

The correct option is D $1$

$\cot \left({123}^0\right)\cot \left({137}^0\right)\cot \left({133}^0\right)\cot \left({147}^0\right)$
$=\cot ({180}^0-{57}^0)\cot ({180}^0-{43}^0)\cot ({180}^0-{47}^0)\cot ({180}^0-{33}^0)$
$=\cot \left({57}^0\right)\cot \left({43}^0\right)\cot \left({47}^0\right)\cot \left({33}^0\right)$
Now
$\cot \left(\alpha \right).\cot \left(\beta \right)=1$ when $\alpha +\beta ={90}^0$
Hence
$\cot \left({57}^0\right)\cot \left({43}^0\right)\cot \left({47}^0\right)\cot \left({33}^0\right)$
$=\left[\cot \right({57}^0).\cot ({33}^0\left)\right]\left[\cot \right({43}^0).\cot ({47}^0\left)\right]$
$=1$