2.7 Supplementary Exercise for JEE Main And Advance

Accessible NCERT Class 12 Mathematics Textbook for Blind and Visually Impaired Students facilitated by Professor T K Bansal.

Question 1


\[\tan^{−1} \frac{2}{11}\ +\ \tan^{−1} \frac{7}{24}\ =\ \tan^{−1} \frac{1}{2}\]

Question 2


\[2 \tan^{−1} \frac{1}{2}\ +\ \tan^{−1} \frac{1}{7}\ =\ \tan^{−1} \frac{31}{17}\]

Question 3


\[\tan^{−1} \frac{1}{\sqrt{x^2−1}},\ |x| > 0\]

Question 4


\[\cot \tan^{−1} a\ +\ \cot^{−1} a\]

Question 5


If


\[\sin sin^{−1} (\frac{1}{5}\ +\ \cos^{−1} x)\ =\ 1,\]


then find the value of x

Question 6


If


\[\tan^{−1} \frac{x−1}{x−2}\ +\ \tan^{−1} \frac{x+1}{x+2}\ =\ \frac{π}{4},\]


then find the value of x

Question 7


\[\tan^{−1} \frac{1}{5}\ +\ \tan^{−1} \frac{1}{7}\ +\ \tan^{−1} \frac{1}{3}\ +\ \tan^{−1} \frac{1}{8}\ =\ \frac{π}{4}\]

Question 8


\[\frac{9π}{8}\ −\ \frac{9}{4} \sin^{−1} \frac{1}{3}\ =\ \frac{9}{4})\ sin^{−1} \frac{2√2}{3}\]

Question 9


\[\tan^{−1} \frac{x}{y}\ −\ \tan^{−1} \frac{x−y}{x+y} =\]


(A) π/2


(B) π/3


(C) π/4


(D) (3π)/4

Answer 9


C