Question

Without using trigonometric tables, evaluate:

$\left(i\right)cos{48}^{\circ }-sin{42}^{\circ }$

$\left(ii\right)cosec{31}^{\circ }-sec{59}^{\circ }$

$\left(iii\right)cot{34}^{\circ }-tan{56}^{\circ }$

$\left(iv\right)co{s}^2{13}^{\circ }-si{n}^2{77}^{\circ }$ [4 MARKS]

Answer 1

Solution: 1 Mark each

$\left(i\right)cos{48}^{\circ }-sin{42}^{\circ }=cos({90}^{\circ }-{42}^{\circ })-sin{42}^{\circ }$

$=sin{42}^{\circ }-sin{42}^{\circ }=0$

$[\because cos({90}^{\circ }-\theta )=sin\theta ]$


$\left(ii\right)cosec{31}^{\circ }-sec{59}^{\circ }=cosec({90}^{\circ }-{59}^{\circ })-sec{59}^{\circ }$

$=sec{59}^{\circ }-sec{59}^{\circ }=0$

$[\because cosec({90}^{\circ }-\theta )=sec\theta ]$


$\left(iii\right)cot{34}^{\circ }-tan{56}^{\circ }=cot({90}^{\circ }-{56}^{\circ })-tan{56}^{\circ }$

$=tan{56}^{\circ }-tan{56}^{\circ }=0$

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


$\left(iv\right)co{s}^2{13}^{\circ }-si{n}^2{77}^{\circ }=co{s}^2({90}^{\circ }-{77}^{\circ })-si{n}^2{77}^{\circ }$

$=si{n}^2{77}^{\circ }-si{n}^2{77}^{\circ }=0$

$[\because cos({90}^{\circ }-\theta )=sin\theta ]$