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Question
If acos2θ + bsin2θ = c has α and β as its roots, then prove that tanα + tanβ = `(2b)/(a + c)`.
`["Hint: Use the identities" cos2theta = (1 - tan^2theta)/(1 + tan^2theta) "and" sin2theta = (2tantheta)/(1 + tan^2theta)]`.
Solution
acos2θ + bsin2θ = c
α and β are the roots of the equation.
Using the formula of multiple angles,
We know that,
cos2θ = `(1 - tan^2theta)/(1 + tan^2theta)` and sin2θ = `(2tantheta)/(1 + tan^2theta)`
∴ `a((1 - tan^2theta)/(1 + tan^2theta)) + b((2tantheta)/(1 + tan^2theta)) - c` = 0
We know that,
The sum of roots of a quadratic equation,
ax2 + bx + c = 0 is given by `((-b)/a)`.
Therefore,
tanα + tanβ = `(-2b)/(-(c + a)`
= `(2b)/((c + a))`
Hence, tanα + tanβ = `(2b)/((c + a))`
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