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Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise Multiple Choice Questions Question 29 Maths Textbbok Solution.

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Answer: 7

Hint: Try to solve \cot ^{-1} function, then move to cot function.

Given: \cot \left(\frac{\pi}{4}-2 \cot ^{-1} 3\right)


Let 2 \cot ^{-1} 3=y

Then\cot \frac{y}{2}=3

\cot \left(\frac{\pi}{4}-2 \cot ^{-1} 3\right)=\cot \left(\frac{\pi}{4}-y\right)

                                        =\frac{\cot \frac{\pi}{4} \cot y+1}{\cot y-\cot \frac{\pi}{4}}

                                        =\frac{\cot y+1}{\cot y-1}

                                        =\frac{\frac{\cot ^{2} \frac{y}{2}-1}{2 \cot \frac{y}{2}}+1}{\frac{\cot ^{2} \frac{y}{2}-1}{2 \cot \frac{y}{2}}-1}

                                        =\frac{\cot ^{2} \frac{y}{2}+2 \cot \frac{y}{2}-1}{\cot ^{2} \frac{y}{2}-2 \cot \frac{y}{2}-1}

                                        \begin{aligned} &=\frac{9+6-1}{9-6-1} \\ &=7 \end{aligned}

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