**Question:**

**If a cos2θ + b sin 2θ = c has α and β as its roots, then prove that tan α + tan β = 2b/(a + c)**

**[Hint: Use the identities cos 2θ = (( 1 – tan2 θ)/(1 + tan2 θ) and sin 2θ = 2tan θ/(1 + tan2 θ)]**

**Solution:**

According to the question,

a cos2θ + b sin 2θ = c

α and β are the roots of the equation.

Using the formula of multiple angles,

We know that,

$\cos 2 \theta=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$ and $\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$

$\therefore \mathrm{a}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)+\mathrm{b}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)-\mathrm{c}=0$

⇒ a(1 – tan2θ) + 2b tan θ – c(1 + tan2θ) = 0

⇒ (-c – a)tan2θ + 2b tan θ – c + a = 0 …(i)

We know that,

The sum of roots of a quadratic equation, ax2 + bx + c = 0 is given by (-b/a)

Therefore,

tan α + tan β = –2b/–(c + a) = 2b/(c + a)

Hence, tan α + tan β = 2b/(c + a)