MHT CET · Maths · Trigonometric Ratios & Identities
If \(A, B, C\) are angles of a \(\triangle A B C\), then \(\tan 2 A+\tan 2 B+\tan 2 C=\)
- A \(\tan 2 A \tan 3 B \tan 2 C\)
- B \(\tan 2 A \tan 2 B \tan 2 C\)
- C \(\tan A \tan B \tan C\)
- D \(\tan 3 A \tan 2 B \tan 2 C\)
Answer & Solution
Correct Answer
(B) \(\tan 2 A \tan 2 B \tan 2 C\)
Step-by-step Solution
Detailed explanation
\( \text { In } \Delta A B C, A+B+C=\pi \Rightarrow 2 A+2 B+2 C=2 \pi \)
\( \therefore 2 A+2 B=2 \pi-2 C \Rightarrow \tan (2 A+2 B)\)\(=\tan (2 \pi-2 C)=-\tan 2 C \)
\( \frac{\tan 2 A+\tan 2 B}{1-\tan 2 A \tan 2 B}=-\tan 2 C \)
\( \therefore \tan 2 A+\tan 2 B=-\tan 2 C(1-\tan 2 A \tan 2 B) \)
\( \therefore \tan 2 A+\tan 2 B+\tan 2 C=\tan 2 A \tan 2 B \tan 2 C\)
\( \therefore 2 A+2 B=2 \pi-2 C \Rightarrow \tan (2 A+2 B)\)\(=\tan (2 \pi-2 C)=-\tan 2 C \)
\( \frac{\tan 2 A+\tan 2 B}{1-\tan 2 A \tan 2 B}=-\tan 2 C \)
\( \therefore \tan 2 A+\tan 2 B=-\tan 2 C(1-\tan 2 A \tan 2 B) \)
\( \therefore \tan 2 A+\tan 2 B+\tan 2 C=\tan 2 A \tan 2 B \tan 2 C\)
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