Inverse Trigonometric Functions
Free NCERT & Exemplar step-by-step solutions — CBSE Class 12 Mathematics
NCERT Exemplar
${\cos ^{ - 1}}\left( {\frac{3}{5}\cos x + \frac{4}{5}\sin x} \right)$, where $x \in \left[ {\frac{{ - 3\pi }}{4},\frac{\pi }{4}} \right]$. LA Q14 Prove that ${\sin ^{ - 1}}\frac{8}{{17}} + {\sin ^{ - 1}}\frac{3}{5} = {\sin ^{ - 1}}\frac{{77}}{{85}}$. LA Q15 Show that ${\sin ^{ - 1}}\frac{5}{{13}} + {\cos ^{ - 1}}\frac{3}{5} = {\tan ^{ - 1}}\frac{{63}}{{16}}$. LA Q16 Prove that ${\tan ^{ - 1}}\frac{1}{4} + {\tan ^{ - 1}}\frac{2}{9} = {\sin ^{ - 1}}\frac{1}{{\sqrt 5 }}$. LA Q17 Find the value of $4{\tan ^{ - 1}}\frac{1}{5} - {\tan ^{ - 1}}\frac{1}{{239}}$. LA Q18 Show that $\tan \left( {\frac{1}{2}{{\sin }^{ - 1}}\frac{3}{4}} \right) = \frac{{4 - \sqrt 7 }}{3}$ and justify why the other value $\frac{{4 + \sqrt 7 }}{3}$ is ignored? LA Q19 If ${a_1},{a_2},{a_3}, \ldots ,{a_n}$ is an arithmetic progression with common difference
$d$, then evaluate the following expression.
$\tan \left[ {{{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_1}{a_2}}}} \right) + {{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_2}{a_3}}}} \right) + {{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_3}{a_4}}}} \right)} \right.$ $\left. { + \ldots + {{\tan }^{ - 1}}\left( {\frac{d}{{1 + {a_{n - 1}}{a_n}}}} \right)} \right]$
We have, ${a_1} = a,{a_2} = a + d,{a_3} = a + 2d$
and $d = {a_2} - {a_1} = {a_3} - {a_2} = {a_4} - {a_3} = \ldots = {a_n} - {a_{n - 1}}$
${\tan ^{ - 1}}\sqrt {x(x + 1)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = \frac{\pi }{2}$. SA Q8 Find the value of $\sin \left( {2{{\tan }^{ - 1}}\frac{1}{3}} \right) + \cos \left( {{{\tan }^{ - 1}}2\sqrt 2 } \right)$. SA Q9 If $2{\tan ^{ - 1}}(\cos \theta ) = {\tan ^{ - 1}}(2{\mathop{\rm cosec}\nolimits} \theta )$, then show that $\theta = \frac{\pi }{4}$, where $n$ is any integer. SA Q10 Show that $\cos \left( {2{{\tan }^{ - 1}}\frac{1}{7}} \right) = \sin \left( {4{{\tan }^{ - 1}}\frac{1}{3}} \right)$. SA Q11 Solve the equation $\cos \left( {{{\tan }^{ - 1}}x} \right) = \sin \left( {{{\cot }^{ - 1}}\frac{3}{4}} \right)$. SA
Exercise 2.1
(A) $0 \le y \le \pi$
(B) $- \frac{\pi }{2} \le y \le \frac{\pi }{2}$
(C) $0 < y < \pi$
(D) $- \frac{\pi }{2} < y < \frac{\pi }{2}$
SA Q14 ${\tan ^{ - 1}}\sqrt 3 - {\sec ^{ - 1}}( - 2)$ is equal to(A) $\pi$
(B) $- \frac{\pi }{3}$
(C) $\frac{\pi }{3}$
(D) $\frac{{2\pi }}{3}$
SAExercise 2.2
(A) $\frac{{7\pi }}{6}$
(B) $\frac{{5\pi }}{6}$
(C) $\frac{\pi }{3}$
(D) $\frac{\pi }{6}$
SA Q20 $\sin \left( {\frac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \frac{1}{2}} \right)} \right)$ is equal to(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{1}{4}$
(D) 1
SA Q21 ${\tan ^{ - 1}}\sqrt 3 - {\cot ^{ - 1}}( - \sqrt 3 )$ is equal to(A) $\pi$
(B) $- \frac{\pi }{2}$
(C) 0
(D) $2\sqrt 3$
SAMiscellaneous Exercise
[Hint : Put $x = \cos 2\theta ]$
SA Q12 $\frac{{9\pi }}{8} - \frac{9}{4}{\sin ^{ - 1}}\frac{1}{3} = \frac{9}{4}{\sin ^{ - 1}}\frac{{2\sqrt 2 }}{3}$ SA Q13 $2{\tan ^{ - 1}}(\cos x) = {\tan ^{ - 1}}(2{\rm{cosec }}x)$ SA Q14 ${\tan ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right) = \frac{1}{2}{\tan ^{ - 1}}x,\;\;(x > 0)$ SA Q15 $\sin ({\tan ^{ - 1}}x),\;|x|\; < 1$ is equal to(A) $\frac{x}{{\sqrt {1 - {x^2}} }}$
(B) $\frac{1}{{\sqrt {1 - {x^2}} }}$
(C) $\frac{1}{{\sqrt {1 + {x^2}} }}$
(D) $\frac{x}{{\sqrt {1 + {x^2}} }}$
SA Q16 . If ${\sin ^{ - 1}}(1 - x) - 2{\sin ^{ - 1}}x = \frac{\pi }{2},\;\;then\;\;x$ is equal to(A) $0,\;\;\frac{1}{2}$
(B) $1,\;\;\frac{1}{2}$
(C) 0
(D) $\frac{1}{2}$
SA Q17 ${\tan ^{ - 1}}\left( {\frac{x}{y}} \right) - {\tan ^{ - 1}}\left( {\frac{{x - y}}{{x + y}}} \right)$ is equal to(A) $\frac{\pi }{2}$
(B) $\frac{\pi }{3}$
(C) $\frac{\pi }{4}$
(D) $\frac{{ - 3\pi }}{4}$
SA