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The equation \( a \cos 2\theta + b \sin2\theta = c\) has 2 roots (of \(\theta\)): \(\alpha\) and \(\beta\). Prove that \( \tan(\alpha + \beta) = \dfrac ba\).

Note by Pritthijit Nath 1 year, 10 months ago

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I think the answer should be a/b instead of b/a. !

Proceed as follows:-

Let the eq.n be :→ a(cos2p) + b(sin2p) = c (just for convenience in writing)

Now we substitute,

x = tan(p)

So that,

Sin2p = 2x/(1+x^2) &

Cos2p = (1-x^2)/(1+x^2)

Now our eq.n reduces to→

a [2x/(1+x^2)] + b [(1-x^2)/(1+x^2)] =c

On further simplification u get a quadratic eq. in x as follows→

(b+c)•x^2 - 2ax + c-b = 0

Now let the two roots be tanA & tanB,

Then by vieta's formulae,

TanA + TanB = 2a/(b+c) &

TanA●TanB = (c-b)/(c+b)

Now use the formula of Tan (A+B)→

Tan (A+B) = (TanA + TanB)/(1-TanA●TanB)

To get→

| Tan(A+B) = a/b |★ ( Ans.)

Please feel free to correct and suggest me wherever im wrong ...

Thank you.!

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TopNewestI think the answer should be a/b instead of b/a. !

Proceed as follows:-

Let the eq.n be :→ a(cos2p) + b(sin2p) = c (just for convenience in writing)

Now we substitute,

x = tan(p)

So that,

Sin2p = 2x/(1+x^2) &

Cos2p = (1-x^2)/(1+x^2)

Now our eq.n reduces to→

a [2x/(1+x^2)] + b [(1-x^2)/(1+x^2)] =c

On further simplification u get a quadratic eq. in x as follows→

(b+c)•x^2 - 2ax + c-b = 0

Now let the two roots be tanA & tanB,

Then by vieta's formulae,

TanA + TanB = 2a/(b+c) &

TanA●TanB = (c-b)/(c+b)

Now use the formula of Tan (A+B)→

Tan (A+B) = (TanA + TanB)/(1-TanA●TanB)

To get→

| Tan(A+B) = a/b |★ ( Ans.)

Please feel free to correct and suggest me wherever im wrong ...

Thank you.!

Log in to reply