TS EAMCET · Maths · Trigonometric Equations
Assertion (A) If \(\alpha=12^{\circ}, \beta=15^{\circ}, \gamma=18^{\circ}\), then \(\tan 2 \alpha \tan 2 \beta+\tan 2 \beta \tan 2 \gamma+\tan 2 \gamma \tan 2 \alpha=1\) Reason (R) In \(\triangle A B C \cdot \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}\) \[ +\tan \frac{C}{2} \tan \frac{A}{2}=1 \] Which of the following is true?
- A Both (A) and (R) are true and (R) is the correct explanation of (A)
- B Both (A) and (R) are true, but (R) is not the correct explanation of (A)
- C (A) is true, but ( \(R\) ) is false
- D (A) is false, but (R) is true
Answer & Solution
Correct Answer
(A) Both (A) and (R) are true and (R) is the correct explanation of (A)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } \alpha=12^{\circ}, \beta=15^{\circ}, \gamma=18^{\circ} \\ & \Rightarrow \tan 2 \alpha \tan 2 \beta+\tan 2 \beta \tan 2 \gamma \\ & +\tan 2 \gamma \tan 2 \alpha=1 \\ & \Rightarrow \tan 2 \alpha[\tan 2 \beta+\tan 2 \gamma]=1-\tan 2 \beta \tan 2 \gamma…
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