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NDA 2011
Exam
NDA
Year
2011
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NDA 2011 · Mathematics · Inverse Trigonometric Functions

In a triangle A B C , if A= ^ -1 2 and B= ^ -1 3 , then C is equal...

In a triangle ABCA B C, if A=tan⁡−12A=\tan ^{-1} 2 and B=tan⁡−13B=\tan ^{-1} 3, then CC is equal to

Options

  1. Aπ3\frac{\pi}{3}
  2. Bπ4\frac{\pi}{4}
  3. Cπ6\frac{\pi}{6}
  4. Dπ2\frac{\pi}{2}

Correct answer

B. π4\frac{\pi}{4}

Step-by-step solution

tan⁡A=2 and tan⁡B=3tan⁡(A+B)==tan⁡A+tan⁡B1−tan⁡A⋅tan⁡B=2+31−6=5−5=−1∵A+B+C=πA+B=π−Ctan⁡(A+B)=tan⁡(π−C)=−tan⁡C∴−tan⁡C=−1⇒tan⁡C=tan⁡π4∴C=π4.\begin{aligned} & \tan \mathrm{A}=2 \text{ and } \tan \mathrm{B}=3 \\ & \qquad \begin{array}{c}\tan(\mathrm{A}+\mathrm{B})= \\ =\frac{\tan \mathrm{A}+\tan \mathrm{B}}{1-\tan \mathrm{A} \cdot \tan \mathrm{B}}=\frac{2+3}{1-6} \\ =\frac{5}{-5}=-1 \\ \because \quad \mathrm{ A}+\mathrm{B}+\mathrm{C}=\pi \\ \mathrm{A}+\mathrm{B}=\pi-\mathrm{C}\end{array} \\ & \tan (\mathrm{A}+\mathrm{B})=\tan (\pi-\mathrm{C})=-\tan \mathrm{C} \\ & \therefore-\tan \mathrm{C}=-1 \Rightarrow \tan \mathrm{C}=\tan \frac{\pi}{4} \\ & \therefore \mathrm{C}=\frac{\pi}{4}.\end{aligned}
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