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If acos2θ + bsin2θ = c has α and β as its roots, then prove that tanα + tanβ = 2ba+c. Hint: Use the identitiesand[Hint: Use the identitiescos2θ=1-tan2θ1+tan2θandsin2θ= 2tanθ1+tan2θ].

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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)]`.

Theorem
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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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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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Chapter 3: Trigonometric Functions - Exercise [Page 54]

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NCERT Exemplar Mathematics Exemplar [English] Class 11
Chapter 3 Trigonometric Functions
Exercise | Q 23 | Page 54

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