NDA 2024
Exam NDA
Year 2024
Subject Mathematics
Chapter Trigonometric Ratios & Identities NDA 2024 · Mathematics · Trigonometric Ratios & Identities
In a triangle ABC, A+ B+ C=k What is the value of A B C ?
In a triangle
A B C , tan A + tan B + tan C = k \mathrm{ABC}, \tan \mathrm{A}+\tan \mathrm{B}+\tan \mathrm{C}=k ABC , tan A + tan B + tan C = k What is the value of
cot A cot B cot C \cot A \cot B \cot C cot A cot B cot C ?
Options A 0.5 k B 1 k \frac{1}{k} k 1 C 3 k \frac{3}{k} k 3 D 1 k 3 \frac{1}{k^3} k 3 1 Correct answer B. 1 k \frac{1}{k} k 1
Step-by-step solution
In
△ A B C , A + B + C = π \triangle \mathrm{ABC}, \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi △ ABC , A + B + C = π
Taking
tan \tan tan both side, we get
⇒ tan ( A + B + C ) = tan π = 0 ⇒ tan A + tan B + tan C − tan A tan B tan C 1 − tan A ⋅ tan B − tan B ⋅ tan C − tan C ⋅ tan A = 0 ⇒ tan A ⋅ tan B ⋅ tan C = tan A + tan B + tan C ⇒ cot A ⋅ cot B ⋅ cot C = 1 tan A + tan B + tan C = 1 K \begin{aligned}
& \Rightarrow \tan (A+B+C)=\tan \pi=0 \\
& \Rightarrow \frac{\tan A+\tan B+\tan C-\tan A \tan B \tan C}{1-\tan A \cdot \tan B-\tan B \cdot \tan C-\tan C \cdot \tan A}=0 \\
& \Rightarrow \tan A \cdot \tan B \cdot \tan C=\tan A+\tan B+\tan C \\
& \Rightarrow \cot A \cdot \cot B \cdot \cot C=\frac{1}{\tan A+\tan B+\tan C} \\
& =\frac{1}{K}
\end{aligned} ⇒ tan ( A + B + C ) = tan π = 0 ⇒ 1 − tan A ⋅ tan B − tan B ⋅ tan C − tan C ⋅ tan A tan A + tan B + tan C − tan A tan B tan C = 0 ⇒ tan A ⋅ tan B ⋅ tan C = tan A + tan B + tan C ⇒ cot A ⋅ cot B ⋅ cot C = tan A + tan B + tan C 1 = K 1
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