Question

if m tan (theta - 30 degree ) = ntan ( theta plus 120 degree ) show that cos 2 theta = m+n divided by 2(m-n)

Answer 1

apply the process of componendo and dividendo

Answer 2

Answer:

If $mtan(\theta-30\degree)=ntan(\theta+120\degree)$

Show that :  $cos2\theta=\frac{m+n}{2(m-n)}$

We take right hand side and will show equal to left hand side

R.H.S

$\frac{m+n}{2(m-n)}$

Since, $m tan(\theta-30\degree)=n tan(\theta+120\degree)$

divide both the sides by $ntan(\theta-30\degree)$

$\frac{m}{n}=\frac{ntan(\theta+120\degree)}{ntan(\theta-30\degree)}$

$\frac{m}{n}=\frac{tan(\theta+120\degree)}{tan(\theta-30\degree)}$

Let $\theta+120=A$ and $\theta-30=B$

$\frac{m}{n}=\frac{tanA}{tanB}$

comparing both sides we get value of m and n

$m=tanA$ and $n=tanB$

put the value of m  and n in $\frac{m+n}{2(m-n)}$

$\frac{tanA+tanB}{2(tanA-tanB)}$

$\frac{\frac{sinA}{cosA}+\frac{sinB}{cosB}}{2(\frac{sinA}{cosA}-\frac{sinB}{cosB})}$

$\frac{sinAcosB+sinBcosA}{2(cosBsinA-sinBcosA)}$

$\frac{sin(A+B)}{2(sin(A-B))}$

$\frac{sin(\theta+120-\theta-30)}{2(sin(\theta+120-\theta+30))}$

$\frac{sin(90+2\theta)}{2(sin(150))}$

$\frac{cos2\theta}{2(sin(180-30))}$

$\frac{cos2\theta}{2(sin30)}$

$\frac{cos2\theta}{2(\frac{1}{2})}$

$cos2\theta$

=R.H.S

Hence proved