If - - be the roots of the equation 3cos 2 - -4sin 2 - - 5- then match the following from List I to List II- List I List II- A tan- -tan- P 0- B tan - - Q 4 3- C tan - - R 1 4- D tan-tan- S 1- -

We know that α, β are the roots of 3cos 2 θ +4sin 2 θ = 5 ⇒ 3(1 tan^2θ1+tan^2θ)+4(2tanθ1+tan^2θ) = 5 ⇒3 3tan^2θ +8tanθ1+tan^2θ = 5 ⇒ nbsp;3 3tan^2θ +8tanθ ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

If α, β be th...

Question

If $α, β$ be the roots of the equation $3 cos 2 θ + 4 sin 2 θ = 5$ , then match the following from $List I$ to $List II$ .

$\begin{matrix} List I & List II (A) & tan α + tan β & (P) & 0 (B) & tan (α + β) & (Q) & \frac{4}{3} (C) & tan (α - β) & (R) & \frac{1}{4} (D) & tan α tan β & (S) & 1 \end{matrix}$

(A) \to (P) (B) \to (Q) (C) \to (R) (D) \to (S)

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(A) \to (P) (B) \to (R) (C) \to (Q) (D) \to (S)

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(A) \to (Q) (B) \to (P) (C) \to (S) (D) \to (R)

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(A) \to (S) (B) \to (Q) (C) \to (P) (D) \to (R)

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Solution

The correct option is D $(A) \to (S) (B) \to (Q) (C) \to (P) (D) \to (R)$
We know that $α, β$ are the roots of
$3 cos 2 θ + 4 sin 2 θ = 5 \Rightarrow 3 (\frac{1 - {tan}^{2} θ}{1 + {tan}^{2} θ}) + 4 (\frac{2 tan θ}{1 + {tan}^{2} θ}) = 5 \Rightarrow \frac{3 - 3 {tan}^{2} θ + 8 tan θ}{1 + {tan}^{2} θ} = 5 \Rightarrow 3 - 3 {tan}^{2} θ + 8 tan θ = 5 + 5 {tan}^{2} θ \Rightarrow 4 {tan}^{2} θ - 4 tan θ + 1 = 0 \dots (1)$
The roots of equation $(1)$ is $tan α, tan β$ ,
Now,
$\Rightarrow (2 tan θ - 1)^{2} = 0 ∴ tan α = tan β = \frac{1}{2}$
Now,
$tan α + tan β = 1 tan α tan β = \frac{1}{4} tan (α - β) = 0 tan (α + β) = \frac{tan α + tan β}{1 - tan α tan β} = \frac{4}{3}$
Therefore,
$(A) \to (S) (B) \to (Q) (C) \to (P) (D) \to (R)$

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If α,β be the roots of the equation 3cos2θ+4sin2θ=5, then match the following from List I to List II. List IList II (A)tanα+tanβ(P)0(B)tan(α+β)(Q)43(C)tan(α−β)(R)14(D)tanαtanβ(S)1