NDA 2024
Exam NDA
Year 2024
Subject Mathematics
Chapter Inverse Trigonometric Functions NDA 2024 · Mathematics · Inverse Trigonometric Functions
What is 15+ ^2 ((π)/(4)-2 ^ -1 3) equal to?
What is \sqrt{15+\cot ^2\left(\frac{\pi}{4}-2 \cot ^{-1} 3\right)} equal to?
Step-by-step solution
2 cot − 1 3 = 2 tan − 1 1 3 = tan − 1 2 3 1 − 1 9 = tan − 1 ( 2 3 × 9 8 ) = tan − 1 3 4 \begin{aligned} 2 \cot ^{-1} 3 & =2 \tan ^{-1} \frac{1}{3}=\tan ^{-1} \frac{\frac{2}{3}}{1-\frac{1}{9}} \\ & =\tan ^{-1}\left(\frac{2}{3} \times \frac{9}{8}\right)=\tan ^{-1} \frac{3}{4}\end{aligned} 2 cot − 1 3 = 2 tan − 1 3 1 = tan − 1 1 − 9 1 3 2 = tan − 1 ( 3 2 × 8 9 ) = tan − 1 4 3
So,
15 + cot 2 ( π 4 − 2 cot − 1 3 ) = 15 + cot 2 ( tan − 1 1 − tan − 1 3 4 ) = 15 + cot 2 ( tan − 1 ( 1 − 3 4 ) 1 + 3 4 ) ] = 15 + cot 2 ( tan − 1 1 7 ) = 15 + 7 2 = 15 + cot 2 ( cot − 1 7 ) = 15 + 49 = 8 \begin{aligned}
\sqrt{15+\cot ^2\left(\frac{\pi}{4}\right.}- & \left.2 \cot ^{-1} 3\right) \\
& =\sqrt{15+\cot ^2\left(\tan ^{-1} 1-\tan ^{-1} \frac{3}{4}\right)} \\
& =\sqrt{15+\cot ^2\left(\tan ^{-1}\left(\frac{\left.1-\frac{3}{4}\right)}{1+\frac{3}{4}}\right)\right]} \\
& =\sqrt{15+\cot ^2\left(\tan ^{-1} \frac{1}{7}\right)} \\
& =\sqrt{15+7^2} \\
& =\sqrt{15+\cot ^2\left(\cot ^{-1} 7\right)} \\
& =\sqrt{15+49}=8
\end{aligned} 15 + cot 2 ( 4 π − 2 cot − 1 3 ) = 15 + cot 2 ( tan − 1 1 − tan − 1 4 3 ) = 15 + cot 2 ( tan − 1 ( 1 + 4 3 1 − 4 3 ) ) ] = 15 + cot 2 ( tan − 1 7 1 ) = 15 + 7 2 = 15 + cot 2 ( cot − 1 7 ) = 15 + 49 = 8
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