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Class 12 Chapter 2 Maths Extra Questions
Chapter 2 Class 12 Mathematics Practice Questions
Chapter 2 Inverse Trigonometric Functions
The period of the function f(x) = cos4x + tan3x is
- {tex}\frac{\pi }{3}{/tex}
- {tex}\pi {/tex}
- None of these
- {tex}\frac{\pi }{2}{/tex}
If {tex}3{\sin ^{ – 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) {/tex} {tex}- 4{\cos ^{ – 1}}\left( {\frac{{1 – {x^2}}}{{1 + {x^2}}}} \right) {/tex}{tex}+ 2{\tan ^{ – 1}}\left( {\frac{{2x}}{{1 – {x^2}}}} \right) = \frac{\pi }{3}{/tex}. Then, x=.
- {tex}\frac{1}{{\sqrt 3 }}{/tex}
- {tex}\frac{1}{{\sqrt 2 }}{/tex}
- 2
- 1
The value of {tex}\tan {15^0} + \cot {15^0}{/tex}is
- 4
- Not defined
- {tex}\sqrt 3 {/tex}
- {tex}2\sqrt 3 {/tex}
The values of x which satisfy the trigonometric equation {tex}{\tan ^{ – 1}}\left( {\frac{{x – 1}}{{x – 2}}} \right) + {\tan ^{ – 1}}\left( {\frac{{x + 1}}{{x + 2}}} \right) = \frac{\pi }{4}{/tex} are:
- {tex} \pm 2{/tex}
- {tex} \pm \frac{1}{2}{/tex}
- {tex} \pm \frac{1}{{\sqrt 2 }}{/tex}
- {tex} \pm \sqrt 2 {/tex}
The minimum value of sinx – cosx is
- {tex} – \sqrt 2 {/tex}
- -1
- 0
- 1
- The principle value of tan-1{tex}\sqrt3{/tex} is ________.
- If y = 2 tan-1x + sin-1{tex}\left(\frac{2x}{1+x^2}\right){/tex} for all x, then ________ < y < ________.
- The value of cos (sin-1x + cos-1x), |x| {tex}\leq{/tex} 1 is ________.
Find the principal value of {tex}{\sin ^{ – 1}}\left( {\frac{1}{{\sqrt 2 }}} \right){/tex}.
Write the principal value of cos{tex}^{-1}{/tex}1 [cos(680)°].
Prove that {tex}{\tan ^{ – 1}}\sqrt x = \frac{1}{2}{\cos ^{ – 1}}\left( {\frac{{1 – x}}{{1 + x}}} \right){/tex}. (1)
Find the value of the expression {tex}{\tan ^{ – 1}}\left( {\tan \frac{{3\pi }}{4}} \right){/tex}.
Solve the equation: 2tan-1(cosx) = tan-1(2cosec x).
Find the value of {tex}{\sin ^{ – 1}}\left( {\sin \frac{{2\pi }}{3}} \right){/tex}. (2)
Prove that {tex} \tan ^ { – 1 } ( 1 ) + \tan ^ { – 1 } ( 2 ) + \tan ^ { – 1 } ( 3 ) = \pi.{/tex}
Solve for x, {tex} \tan ^ { – 1 } \frac { x } { 2 } + \tan ^ { – 1 } \frac { x } { 3 } = \frac { \pi } { 4 } , \sqrt { 6 } > x > 0.{/tex}
Find the value of the following: {tex}{\tan ^{ – 1}}\left[ {2\cos \left( {2{{\sin }^{ – 1}}\frac{1}{2}} \right)} \right]{/tex}.
Show that {tex}{\sin ^{ – 1}}\frac{{12}}{{13}} + {\cos ^{ – 1}}\frac{4}{5} + {\tan ^{ – 1}}\frac{{63}}{{16}} = \pi{/tex}.
CBSE Test Paper 01
Chapter 2 Inverse Trigonometric Functions
Solution
- {tex}\pi {/tex}, Explanation: {tex}f\left( \pi \right) = \left( {cos\;4\pi + tan3\pi \;} \right) {/tex} gives the same value as f (0). Therefore, the period of the function is {tex}\pi {/tex}.
- {tex}\frac{1}{{\sqrt 3 }}{/tex}, Explanation: {tex}3{\sin ^{ – 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) – 4{\cos ^{ – 1}}\left( {\frac{{1 – {x^2}}}{{1 + {x^2}}}} \right) + 2{\tan ^{ – 1}}\left( {\frac{{2x}}{{1 – {x^2}}}} \right) = \frac{\pi }{3}{/tex}
Put x = tan{tex}\theta{/tex}
{tex}3{\sin ^{ – 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right) – 4{\cos ^{ – 1}}\left( {\frac{{1 – {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right) {/tex}{tex}+ 2{\tan ^{ – 1}}\left( {\frac{{2\tan \theta }}{{1 – {{\tan }^2}\theta }}} \right)\,\, = \frac{\pi }{3}{/tex}
{tex}3{\sin ^{ – 1}}(\sin 2\theta )\, – 4{\cos ^{ – 1}}(\cos 2\theta ) + 2{\tan ^{ – 1}}\left( {\tan 2\theta } \right)\, = \frac{\pi }{3}{/tex}
{tex}3.2\theta \, – 4.2\theta \, + 2.2\theta \, = \,\frac{\pi }{3}\,\,\,\, \Rightarrow 2\theta = \frac{\pi }{3}{/tex}{tex}\Rightarrow \theta = \frac{\pi }{6}{/tex}
{tex}\therefore \,\,{\tan ^{ – 1}}x = \frac{\pi }{6}\,\,\,\, \Rightarrow \,\,x = \tan \left( {\frac{\pi }{6}} \right)\,\, = \frac{1}{{\sqrt 3 }}{/tex}
- {tex}\frac{1}{{\sqrt 3 }}{/tex}, Explanation: {tex}3{\sin ^{ – 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right) – 4{\cos ^{ – 1}}\left( {\frac{{1 – {x^2}}}{{1 + {x^2}}}} \right) + 2{\tan ^{ – 1}}\left( {\frac{{2x}}{{1 – {x^2}}}} \right) = \frac{\pi }{3}{/tex}
- 4, Explanation: {tex}\tan {15^0} + \cot {15^0} = \frac{{\sqrt 3 – 1}}{{\sqrt 3 + 1}} + \frac{{\sqrt 3 + 1}}{{\sqrt 3 – 1}}{/tex}
{tex}= \frac{{{{(\sqrt 3 – 1)}^2} + {{(\sqrt 3 + 1)}^2}}}{2}\; {/tex} = {tex} \frac{8}{2} = 4{/tex}
- 4, Explanation: {tex}\tan {15^0} + \cot {15^0} = \frac{{\sqrt 3 – 1}}{{\sqrt 3 + 1}} + \frac{{\sqrt 3 + 1}}{{\sqrt 3 – 1}}{/tex}
- {tex} \pm \frac{1}{{\sqrt 2 }}{/tex}, Explanation: {tex}{\tan ^{ – 1}}\left( {\frac{{x – 1}}{{x – 2}}} \right)\, + \,\,{\tan ^{ – 1}}\left( {\frac{{x + 1}}{{x + 2}}} \right)\,\, = \frac{\pi }{4}{/tex}
{tex}{\tan ^{ – 1}}\left[ {\frac{{\left( {\frac{{x – 1}}{{x – 2}}} \right)\, + \left( {\frac{{x + 1}}{{x + 2}}} \right)}}{{1 – \left( {\frac{{x – 1}}{{x – 2}}} \right)\left( {\frac{{x + 1}}{{x + 2}}} \right)}}} \right]\,\,\,\,\, = \,\frac{\Pi }{4}{/tex}
{tex}{\tan ^{ – 1}}\left[ {\frac{{\left( {x – 1} \right)\,\left( {x + 2} \right) + \left( {x + 1} \right)\left( {x – 2} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right) – \left( {x + 1} \right)\left( {x – 1} \right)}}} \right]\,\,\, = \,\frac{\Pi }{4}{/tex}
{tex}\left( {\frac{{{x^2}\, + x – 2 + {x^2} – x – 2}}{{{x^2}\, – 4 – {x^2} + 1}}} \right)\, = {\tan ^{ – 1}}\left( {\frac{\Pi }{4}} \right){/tex}
{tex}\left( {\frac{{2{x^2} – 4}}{{ – 3}}} \right)\, = \,1{/tex}
{tex}\therefore \,\,2{x^2} – 4 = – 3{/tex}
{tex} \Rightarrow \,\,2{x^2} = \,1{/tex}
{tex}x = \pm \frac{1}{\sqrt 2 }{/tex}
- {tex} \pm \frac{1}{{\sqrt 2 }}{/tex}, Explanation: {tex}{\tan ^{ – 1}}\left( {\frac{{x – 1}}{{x – 2}}} \right)\, + \,\,{\tan ^{ – 1}}\left( {\frac{{x + 1}}{{x + 2}}} \right)\,\, = \frac{\pi }{4}{/tex}
- {tex} – \sqrt 2 {/tex}, Explanation: Since, range of sine function and cosine function is [-1,1]. But, sine is increasing function and cosine is decreasing function. Therefore, the lowest that both together can attain is {tex}-45^0{/tex}.
{tex}\left( { – \frac{1}{{\sqrt 2 }}} \right) + \left( { – \frac{1}{{\sqrt 2 }}} \right) = – \sqrt 2 {/tex}
- {tex} – \sqrt 2 {/tex}, Explanation: Since, range of sine function and cosine function is [-1,1]. But, sine is increasing function and cosine is decreasing function. Therefore, the lowest that both together can attain is {tex}-45^0{/tex}.
- {tex}\frac{\pi}{3}{/tex}
- -2{tex}\pi{/tex}, 2{tex}\pi{/tex}
- 0
- Let {tex}{\sin ^{ – 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) = \theta{/tex}
{tex}\Rightarrow\sin \theta = \frac{1}{{\sqrt 2 }}{/tex}
We know that {tex}\theta \in \left[ {\frac{{ – \pi }}{2},\frac{\pi }{2}} \right]{/tex}
{tex}\Rightarrow \sin \theta = \sin \frac{\pi }{4}{/tex} {tex}\Rightarrow \theta = \frac{\pi }{4}{/tex}
Therefore, principal value of {tex}{\sin ^{ – 1}}\left( {\frac{1}{{\sqrt 2 }}} \right){/tex} is {tex}\frac{\pi }{4}{/tex} - We know that, principal value branch of {tex}\cos ^ { – 1 }{/tex} x is [0, 180°].
Since, 680° {tex}\in{/tex} [0,180°], so write 680° as 2 {tex} \times {/tex} 360°-40°
Now, {tex}\cos ^ { – 1 }{/tex}[cos (680)°] = {tex}\cos ^ { – 1 }{/tex} [cos(2 ;{tex} \times{/tex} 360°-40°)] = {tex}\cos ^ { – 1 }{/tex}(cos40°) {tex}[ \because \cos ( 4 \pi – \theta ) = \cos \theta ]{/tex}
Since, 40°{tex}\in{/tex} [0,180°] {tex}\therefore \cos ^ { – 1 }{/tex}[cos(680°)] = 40°
{tex}\left[ \because \cos ^ { – 1 } ( \cos \theta ) = \theta ; \forall \theta \in \left[ 0,180 ^ { \circ } \right] \right]{/tex}
which is the required principal value. - LHS = {tex}{\tan ^{ – 1}}\sqrt x {/tex}
Let {tex}\\tan \theta = \sqrt x{/tex}
{tex}{\tan ^2}\theta = x{/tex}
R.H.S. {tex} = \frac{1}{2}{\cos ^{ – 1}}\left( {\frac{{1 – {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right){/tex}
{tex}= \frac{1}{2}{\cos ^{ – 1}}\left( {\cos 2\theta } \right) = \frac{1}{2} \times 2\theta = \theta{/tex}
{tex} = {\tan ^{ – 1}}\sqrt x{/tex} - {tex}{\tan ^{ – 1}}\left( {\tan \frac{{3\pi }}{4}} \right){/tex}
{tex} = {\tan ^{ – 1}}\left( {\tan \frac{{4\pi – \pi }}{4}} \right){/tex}
{tex} = {\tan ^{ – 1}}\left[ {\tan \left( {\pi – \frac{\pi }{4}} \right)} \right]{/tex}
{tex} = {\tan ^{ – 1}}\left[ { – \tan \frac{\pi }{4}} \right]{/tex}
{tex} = {\tan ^{ – 1}}\tan \left( { – \frac{\pi }{4}} \right){/tex} {tex} = – \frac{\pi }{4}{/tex} - 2 tan-1(cos x) = tan-1(2 cosec x)
{tex}\Rightarrow {\tan ^{ – 1}}\left( {\frac{{2\cos x}}{{1 – {{\cos }^2}x}}} \right) = {\tan ^{ – 1}}\left( {\frac{2}{{\sin x}}} \right){/tex}
{tex}\Rightarrow \frac{{2\cos x}}{{1 – {{\cos }^2}x}} = \frac{2}{{\sin x}}{/tex}
{tex}\Rightarrow \frac{{\cos x}}{{\sin x}} = 1{/tex}
{tex}\Rightarrow{/tex} cot x = 1 {tex}\Rightarrow x = \frac{\pi }{4}{/tex} - {tex}{\sin ^{ – 1}}\left( {\sin \frac{{2\pi }}{3}} \right){/tex}
{tex}= {\sin ^{ – 1}}\left( {\sin \frac{{3\pi – \pi }}{3}} \right){/tex}
{tex}= {\sin ^{ – 1}}\left[ {\sin \left( {\pi – \frac{\pi }{3}} \right)} \right]{/tex}
{tex}= {\sin ^{ – 1}}\sin \frac{\pi }{3}{/tex} {tex} = \frac{\pi }{3}{/tex} - To prove, {tex} tan^{-1} (1) + tan^{-1} (2) + tan^{-1} (3) = \pi{/tex}
LHS = {tex}tan^{-1} (1) + tan^{-1} (2) + tan^{-1} (3) {/tex}
{tex} = \tan ^ { – 1 } \left( \tan \frac { \pi } { 4 } \right) + \frac { \pi } { 2 } – \cot ^ { – 1 } ( 2 ) + \frac { \pi } { 2 } – \cot ^ { – 1 } ( 3 ){/tex} {tex} \left[ \because \tan ^ { – 1 } x + \cot ^ { – 1 } x = \frac { \pi } { 2 } \right]{/tex}
{tex} = \frac { \pi } { 4 } + \pi – \left[ \cot ^ { – 1 } ( 2 ) + \cot ^ { – 1 } ( 3 ) \right]{/tex}{tex} \left[ \because \tan ^ { – 1 } ( \tan \theta ) = \theta ; \forall \theta \in \left( – \frac { \pi } { 2 } , \frac { \pi } { 2 } \right) \right]{/tex}
{tex} = \frac { 5 \pi } { 4 } – \left[ \tan ^ { – 1 } \left( \frac { 1 } { 2 } \right) + \tan ^ { – 1 } \left( \frac { 1 } { 3 } \right) \right]{/tex}{tex} \left[ \because \cot ^ { – 1 } x = \tan ^ { – 1 } \frac { 1 } { x } , x > 0 \right]{/tex}
{tex} = \frac { 5 \pi } { 4 } – \left[ \tan ^ { – 1 } \left( \frac { \frac { 1 } { 2 } + \frac { 1 } { 3 } } { 1 – \frac { 1 } { 2 } \cdot \frac { 1 } { 3 } } \right) \right]{/tex}{tex} \left[ \because \tan ^ { – 1 } x + \tan ^ { – 1 } y = \tan ^ { – 1 } \left( \frac { x + y } { 1 – x y } \right), \text { if } x y < 1 \right]{/tex}
{tex} = \frac { 5 \pi } { 4 } – \tan ^ { – 1 } \left( \frac { 5 / 6 } { 5 / 6 } \right){/tex}
{tex} = \frac { 5 \pi } { 4 } – \tan ^ { – 1 } ( 1 ) = \frac { 5 \pi } { 4 } – \frac { \pi } { 4 } = \frac { 4 \pi } { 4 } = \pi{/tex}= RHS (Hence Proved) - Here, we have to find the value of x .Now, we are given that
{tex} \tan ^ { – 1 } \frac { x } { 2 } + \tan ^ { – 1 } \frac { x } { 3 } {/tex}{tex}= \frac { \pi } { 4 } , \sqrt { 6 } > x > 0 {/tex}
{tex}\Rightarrow \tan ^ { – 1 } \left( \frac { \frac { x } { 2 } + \frac { x } { 3 } } { 1 – \frac { x ^ { 2 } } { 6 } } \right) = \frac { \pi } { 4 }{/tex}{tex} \left[ \because \tan ^ { – 1 } x + \tan ^ { – 1 } y = \tan ^ { – 1 } \left( \frac { x + y } { 1 – x y } \right) ; x y < 1 \right]{/tex}
{tex} \Rightarrow \quad \frac { \frac { 3 x + 2 x } { 6 } } { \frac { 6 – x ^ { 2 } } { 6 } } = \tan \frac { \pi } { 4 }{/tex} { taking tan on both sides}
{tex}\Rightarrow \frac { 5 x } { 6 – x ^ { 2 } } = 1 \left[ \because \tan \frac { \pi } { 4 } = 1 \right]{/tex}
{tex} \Rightarrow{/tex} 5x = 6-x2
{tex}\Rightarrow{/tex} x2 + 5x – 6 = 0
{tex}\Rightarrow{/tex} x2 + 6x – x – 6 = 0
{tex}\Rightarrow{/tex} x (x + 6) – 1 (x + 6) = 0
{tex}\Rightarrow{/tex} (x-1) (x + 6) = 0
{tex}\therefore{/tex} x = 1 or – 6
But it is given that, {tex} \,\sqrt {\text{6}} {\text{ > x > 0 }} \Rightarrow {\text{x > 0}}{/tex}
{tex} \therefore{/tex} x = – 6 is rejected.
Hence, x = 1 is the only solution of the given equation. - {tex}{\tan ^{ – 1}}\left[ {2\cos \left( {2{{\sin }^{ – 1}}\frac{1}{2}} \right)} \right]{/tex}
{tex} = {\tan ^{ – 1}}\left[ {2\cos \left( {2{{\sin }^{ – 1}}\sin \frac{\pi }{6}} \right)} \right]{/tex}
{tex} = {\tan ^{ – 1}}\left[ {2\cos \left( {2 \times \frac{\pi }{6}} \right)} \right]{/tex}
{tex} = {\tan ^{ – 1}}\left[ {2\cos \frac{\pi }{3}} \right]{/tex}
{tex}= {\tan ^{ – 1}}\left[ {2 \times \frac{1}{2}} \right]{/tex} = tan-11
{tex}= {\tan ^{ – 1}}\tan \frac{\pi }{4} = \frac{\pi }{4}{/tex} - Let {tex}\theta {/tex} = sin-1({tex}\frac{12}{13}{/tex})
{tex}\Rightarrow{/tex} sin{tex}\theta{/tex} = {tex}\frac{12}{13}{/tex}
{tex}\Rightarrow\sqrt{1-cos^2\theta}=\frac{12}{13}{/tex}
{tex}\Rightarrow 1-cos^2\theta=\frac{(12)^2}{(13)^2}{/tex}
{tex}\Rightarrow cos^2\theta=\frac{(5)^2}{(13)^2}{/tex}
{tex}\Rightarrow cos\theta=\frac{5}{13}{/tex}
Since, tan{tex}\theta{/tex} = {tex}\frac{sin\theta}{cos\theta}=\frac{\frac{12}{13}}{\frac{5}{13}}=\frac{12}{5}{/tex}
{tex}\Rightarrow \theta=tan^{-1}(\frac{12}{5}){/tex}
Thus, {tex}\theta=sin^{-1}(\frac{12}{13})=tan^{-1}(\frac{12}{5}){/tex}
Similarly, {tex}cos^{-1}(\frac{5}{13})=tan^{-1}(\frac{12}{5}){/tex}
We have, LHS = {tex}{\sin ^{ – 1}}\left( {\frac{{12}}{{13}}} \right) + {\cos ^{ – 1}}\left( {\frac{4}{5}} \right) + {\tan ^{ – 1}}\left( {\frac{{63}}{{16}}} \right){/tex}
{tex} = {\tan ^{ – 1}}\left( {\frac{{12}}{5}} \right) + {\tan ^{ – 1}}\left( {\frac{3}{4}} \right) + {\tan ^{ – 1}}\left( {\frac{{63}}{{16}}} \right){/tex}
{tex} = \left[ {{{\tan }^{ – 1}}\left( {\frac{{12}}{5}} \right) + {{\tan }^{ – 1}}\left( {\frac{3}{4}} \right)} \right] + {\tan ^{ – 1}}\left( {\frac{{63}}{{16}}} \right){/tex}
{since {tex}\frac{{12}}{5} \times \frac{3}{4} = \frac{9}{5} > 1,{/tex} therefore , tan-1A + tan-1B = {tex}\pi + {\tan ^{ – 1}}\frac{{A + \_B}}{{1 – AB}}{/tex})
={tex}\pi + {\tan ^{ – 1}}\left( {\frac{{\frac{{12}}{5} + \frac{3}{4}}}{{1 – \left( {\frac{{12}}{5}} \right)\left( {\frac{3}{4}} \right)}}} \right) + {\tan ^{ – 1}}\left( {\frac{{63}}{{16}}} \right){/tex}
= {tex}\pi + {\tan ^{ – 1}}\left( { – \frac{{63}}{{16}}} \right) + {\tan ^{ – 1}}\left( {\frac{{63}}{{16}}} \right){/tex}
{tex} = \pi – {\tan ^{ – 1}}\left( {\frac{{63}}{{16}}} \right) + {\tan ^{ – 1}}\left( {\frac{{63}}{{16}}} \right){/tex} = {tex}\pi{/tex} Hence Proved.
Chapter Wise Important Questions Class 12 Maths Part I and Part II
- Relations and Functions
- Inverse Trigonometric Functions
- Matrices
- Determinants
- Continuity and Differentiability
- Application of Derivatives
- Integrals
- Application of Integrals
- Differential Equations
- Vector Algebra
- Three Dimensional Geometry
- Linear Programming
- Probability
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