In the ABC- BC is produced to D and - ACD- 3-4and tan A- tan B are roots of the equation x2- x- -0- Then

The correct option is C λ =μ 1 tan A+tan B = λ, #160; tan A.tan B = μ Now given ∠ C = π4 ∴ A+B = 3π4 ⇒tan(A+B) = 1 = tan A+tan B1 ...

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Standard XII

Mathematics

Sin(A+B)Sin(A-B)

In the ABC,...

Question

In the $△ A B C$ , $B C$ is produced to D and $∠ A C D = \frac{3 π}{4}$ and $tan A, tan B$ are roots of the equation $x^{2} - λ x + μ = 0.$ Then

λ^{2} - μ^{2} = 1 + 6 μ

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λ^{2} - μ^{2} = 1

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λ = μ - 1

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none of these

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Solution

The correct option is C $λ = μ - 1$
$tan A + tan B = λ, tan A . tan B = μ$
Now given $∠ C = \frac{π}{4}$

$∴ A + B = \frac{3 π}{4}$
$\Rightarrow tan (A + B) = - 1 = \frac{tan A + tan B}{1 - tan A . tan B} = \frac{λ}{1 - μ}$
$\Rightarrow λ = μ - 1$

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In the △ABC, BC is produced to D and ∠ACD=3π4and tanA,tanB are roots of the equation x2−λx+μ=0. Then

The correct option is C λ=μ−1tanA+tanB=λ,tanA.tanB=μ Now given ∠C=π4∴A+B=3π4⇒tan(A+B)=−1=tanA+tanB1−tanA.tanB=λ1−μ⇒λ=μ−1

In the $△ A B C$ , $B C$ is produced to D and $∠ A C D = \frac{3 π}{4}$ and $tan A, tan B$ are roots of the equation $x^{2} - λ x + μ = 0.$ Then

The correct option is C $λ = μ - 1$
$tan A + tan B = λ, tan A . tan B = μ$
Now given $∠ C = \frac{π}{4}$

$∴ A + B = \frac{3 π}{4}$
$\Rightarrow tan (A + B) = - 1 = \frac{tan A + tan B}{1 - tan A . tan B} = \frac{λ}{1 - μ}$
$\Rightarrow λ = μ - 1$