2.5 Miscellaneous Exercise on Chapter 2 - Inverse Trigonometric Functions - Class 12 Mathematics
2.5 Miscellaneous Exercise on Chapter 2
Screen Readable NCERT Class 12 Mathematics Textbook for Blind and Visually Impaired Students prepared by Professor T K Bansal.
Find the value of the following:
Question 1
\[\cos^{−1} (\cos \frac{13π}{6})\]
Answer 1
π/6
Question 2
\[\tan^{−1} (\tan \frac{7π}{6})\]
Answer 2
π/6
Prove that
Question 3
\[2 \sin^{−1} \frac{3}{5}\ =\ \tan^{−1} \frac{24}{7}\]
Question 4
\[\sin^{−1} \frac{8}{17}\ +\ \sin^{−1} \frac{3}{5}\ =\ \tan^{−1} \frac{77}{36}\]
Question 5
\[\cos^{−1} \frac{4}{5}\ +\ \cos^{−1} \frac{12}{13}\ =\ \cos^{−1} \frac{33}{65}\]
Question 6
\[\cos^{−1} \frac{12}{13}\ +\ \sin^{−1} \frac{3}{5}\ =\ \sin^{−1} \frac{56}{65}\]
Question 7
\[\tan^{−1} \frac{63}{16}\ =\ \sin^{−1} \frac{5}{13}\ +\ \cos^{−1} \frac{3}{5}\]
Prove that
Question 8
\[\tan^{−1} √x\ =\ \frac{1}{2}\ \cos^{−1} \frac{1−x}{1+x},\ x\ ∈\ [0,\ 1]\]
Question 9
\[cot^{−1} (\frac{\sqrt{1+\sin x}\ +\ \sqrt{1− \sin x}}{\sqrt{1+\sin x} − \sqrt{1−\sin x}})\ =\ \frac{x}{2},\ x ∈ (0,\ \frac{π}{4})\]
Question 10
\[tan^{−1} (\frac{\sqrt{1+x}\ −\ \sqrt{1−x}}{\sqrt{1+x}\ +\ \sqrt{1−x}})\ =\ \frac{π}{4}\ −\ \frac{1}{2 \cos^{−1} x},\ \frac{−1}{√2} ≤x≤1\]
[Hint: Put x = cos 2 θ]
Solve the following equations:
Question 11
\[2 \tan^{−1} \cos x\ =\ \tan^{−1} (2 \csc x)\]
Answer 11
x = nπ + (π/4), n ∈ Ƶ
Question 12
\[\tan^{−1} \frac{1−x}{1+x}\ =\ \frac{1}{2} \tan^{−1}x,\ x>0 \]
Answer 12
x = 1/√3
Question 13
\[\sin \tan^{−1}x,\ |x|\ <\ 1\ =\]
(A) x/√(1 − x^2)
(B) 1/√(1 − x^2)
(C) 1/√(1 + x^2)
(D) x/√(1 + x^2)
Answer 13
D
Question 14
\[\sin^{−1} (1−x)\ −\ 2\sin^{−1}x\ =\ \frac{π}{2},\]
then x is equal to
(A) 0, ½
(B) 1, ½
(C) 0
(D) ½
Answer 14
C