Question

If $msin\theta =ncos\theta$, then show that:

$\frac{tan\theta +cot\theta }{tan\theta -cot\theta }=\frac{nsin\theta +mcos\theta }{nsin\theta -mcos\theta }=\frac{n^2+m^2}{n^2-m^2}$ [4 MARKS]

Answer 1

Solution: 2 Marks each

$\left(a\right)\frac{tan\theta +cot\theta }{tan\theta -cot\theta }=\frac{\frac{n}{m}+\frac{m}{n}}{\frac{n}{m}-\frac{m}{n}}$

$=\frac{\frac{n^2+m^2}{mn}}{\frac{n^2-m^2}{mn}}$

$=\frac{n^2+m^2}{n^2-m^2}.......\left(1\right)$


$\left(b\right)\frac{nsin\theta +mcos\theta }{nsin\theta -mcos\theta }$

$÷cos\theta$

$\Rightarrow \frac{ntan\theta +m}{ntan\theta +m}$

$=\frac{n\left(\frac{n}{m}\right)+m}{n\left(\frac{n}{m}\right)-m}$

$=\frac{\frac{n^2+m^2}{m}}{\frac{n^2-m^2}{m}}$

$=\frac{n^2+m^2}{n^2-m^2}.......\left(2\right)$

From (1) and (2)

$\frac{tan\theta +cot\theta }{tan\theta -cot\theta }=\frac{nsin\theta +mcos\theta }{nsin\theta -mcos\theta }=\frac{n^2+m^2}{n^2-m^2}$