Question

Without using trigonometric tables, prove that:

$\left(i\right)tan{48}^{\circ }tan{23}^{\circ }tan{42}^{\circ }tan{67}^{\circ }=1$

$\left(ii\right)tan{7}^{\circ }tan{23}^{\circ }tan{60}^{\circ }tan{67}^{\circ }tan{83}^{\circ }=\sqrt{3}$

$\left(iii\right)cot{12}^{\circ }cot{38}^{\circ }cot{52}^{\circ }cot{60}^{\circ }cot{78}^{\circ }=\frac{1}{\sqrt{3}}$ [3 MARKS]

Answer 1

Solution: 1 Mark each

We have:

$\left(i\right)tan{48}^{\circ }tan{23}^{\circ }tan{42}^{\circ }tan{67}^{\circ }$

$=\left(tan{48}^{\circ }tan{42}^{\circ }\right)\left(tan{23}^{\circ }tan{67}^{\circ }\right)$

$=\{tan{48}^{\circ }.tan({90}^{\circ }-{48}^{\circ }\left)\right\}\{tan{23}^{\circ }.tan({90}^{\circ }-{23}^{\circ }\left)\right\}$

$=\left(tan{48}^{\circ }cot{48}^{\circ }\right)\left(tan{23}^{\circ }cot{23}^{\circ }\right)$

$[\because tan({90}^{\circ }-\theta )=cot\theta ]$

$=1\times 1=1$


$\left(ii\right)tan{7}^{\circ }tan{23}^{\circ }tan{60}^{\circ }tan{67}^{\circ }tan{83}^{\circ }$

$=tan{7}^{\circ }tan{83}^{\circ }.tan{23}^{\circ }.tan({90}^{\circ }-{23}^{\circ }).tan{60}^{\circ }$

$=(tan{7}^{\circ }.cot{7}^{\circ }).(tan{23}^{\circ }.cot{23}^{\circ }).tan{60}^{\circ }$

$[\because tan({90}^{\circ }-\theta )=cot\theta ]$

$=1\times 1\times \sqrt{3}=\sqrt{3}$

$[\because tan\theta .cot\theta =1andtan{60}^{\circ }=\sqrt{3}]$


$\left(iii\right)cot{12}^{\circ }cot{38}^{\circ }cot{52}^{\circ }cot{60}^{\circ }cot{78}^{\circ }$

$=(cot{12}^{\circ }.cot{78}^{\circ }).(cot{38}^{\circ }.cot{52}^{\circ }).cot{60}^{\circ }$

$=\{cot{12}^{\circ }.cot({90}^{\circ }-{12}^{\circ }\left)\right\}\{cot{38}^{\circ }.cot({90}^{\circ }-{38}^{\circ }\left)\right\}.cot{60}^{\circ }$

$=(cot{12}^{\circ }.tan{12}^{\circ }).(cot{38}^{\circ }.tan{38}^{\circ }).cot{60}^{\circ }$

$[\because cot({90}^{\circ }-\theta )=tan\theta ]$

$=1\times 1\times \frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}}$ $[\because cot\theta .tan\theta =1andcot{60}^{\circ }=\frac{1}{\sqrt{3}}]$