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CBSE Questions for Class 12 Commerce Maths Inverse Trigonometric Functions Quiz 5 - MCQExams.com
CBSE
Class 12 Commerce Maths
Inverse Trigonometric Functions
Quiz 5
If $$f ( x ) = \sqrt { \sec ^ { - 1 } \left( \frac { 2 - | x | } { 4 } \right) } ,$$ then the domain of $$f ( x )$$ is ______________.
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$$[ - 2,2 ]$$
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$$[ - 6,6 ]$$
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$$( - \infty , - 6 ] \cup [ 6 , \infty )$$
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None Of These
If $$3{\tan ^{ - 1}}\left( {\frac{1}{{2 + \sqrt 3 }}} \right) - {\tan ^{ - 1}}\frac{1}{x} = {\tan ^{ - 1}}\frac{1}{3}$$ then $$x = $$
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0%
$$1$$
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$$2$$
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$$3$$
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$$\sqrt 3 $$
Explanation
Let $$y=\text{tan}^{-1} \bigg(\dfrac{1}{2+\sqrt{3}}\bigg)\implies \tan y=\dfrac{1}{2+\sqrt{3}}$$
$$\tan 3 y=\dfrac{3\tan y-\tan^3 y}{1+3\tan^2 y}=\dfrac{\frac{1}{2+\sqrt{3}}(3-\frac{1}{(2+\sqrt{3})^2})}{1+\frac{3}{(2+\sqrt{3})^2}}=\dfrac{\frac{1}{2+\sqrt{3}}(20+12\sqrt{3})}{4+4\sqrt{3}}=\dfrac{20+12\sqrt{3}}{20+12\sqrt{3}}=1$$
$$\implies 3\text{tan}^{-1}\bigg(\dfrac{1}{2+\sqrt{3}}\bigg)=\text{tan}^{-1}(1)$$
$$\text{tan}^{-1}\bigg(\dfrac{1}{x}\bigg)=\text{tan}^{-1}(1)-\text{tan}^{-1}\dfrac{1}{3}=\text{tan}^{-1}\dfrac{1-\dfrac{1}{3}}{1+\dfrac{1}{3}}=\dfrac{1}{2}$$
$$\implies x=2$$
$$\displaystyle \sum _{ r=1 }^{ n }{ { tan }^{ -1 } } \left(\dfrac { { 2 }^{ r-1 } }{ 1+{ 2 }^{ 2r-1 } } \right)$$ is equal to :
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$${ tan }^{ -1 }({ 2 }^{ n })$$
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$${ tan }^{ -1 }({ 2 }^{ n })-\frac { \pi }{ 4 } $$
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$${ tan }^{ -1 }({ 2 }^{ n+1 })$$
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$${ tan }^{ -1 }({ 2 }^{ n+1 })-\frac { \pi }{ 4 } $$
Explanation
$$\displaystyle\sum ^{n}_{r=1}\text{tan}^{-1}\bigg(\dfrac{2^{r-1}}{1+2^{2 r-1}}\bigg)=\sum^{n}_{r=1}\text{tan}^{-1}\bigg(\dfrac{2^{r-1}(2-1)}{1+2^r.2^{r-1}}\bigg)$$
$$=\displaystyle\sum^{n}_{r=1}\text{tan}^{-1}\bigg(\dfrac{2^r-2^{r-1}}{1+2^r.2^{r-1}}\bigg)$$
$$=\displaystyle\sum^{n}_{r=1}\text{tan}^{-1}{2^r}-\text{tan}^{-1}(2^{r-1})$$
$$=\text{tan}^{-1} 2-\text{tan}^{-1}1+\text{tan}^{-1}2^2-\text{tan}^{-1}2^1+\cdots+\text{tan}^{-1}{2^n}-\text{tan}^{-1}2^{n-1}$$
$$=\text{tan}^{-1}(2^n)-\dfrac{\pi}{4}$$
For $$0<x<1$$ the value of $$\cos^{-1}x+\cos^{-1}(-x)=?$$
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$$0$$
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$$\pi$$
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$$-\pi$$
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none of these
Explanation
We've for $$|x|\le 1$$, $$\cos^{-1} (-x)=\pi -\cos^{-1}x$$.
It also holds in this case.
So for $$0<x<1$$ we've,
$$\cos^{-1}x+\cos^{-1} (-x)$$
$$=\cos^{-1} x+\pi-\cos^{-1}x$$
$$=\pi$$.
The value of $$\sin\ h(\cos\ h^{-1}x)$$ is
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$$\sqrt{x^{2}+1}$$
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$$1/\sqrt{x^{2}+1}$$
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$$\sqrt{x^{2}-1}$$
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$$none of these$$
The value of $$\displaystyle \sum_{r=0}^{\infty}{ tan}^{1}\left(\dfrac{1}{1+r+{r}^{2}}\right)$$ is equal to
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$$\dfrac { \pi } { 4 }$$
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$$- \dfrac { \pi } { 2 }$$
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$$\pi$$
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$$\dfrac { \pi } { 2 }$$
Explanation
given, $$\displaystyle\sum^{\infty}_{r=0}\tan^{-1}\left(\dfrac{1}{1+r+r^2}\right)$$
Consider $$\tan^{-1}\left(\dfrac{1}{1+r+r^2}\right)=\tan^{-1}\left(\dfrac{1}{1+r(r+1)}\right)$$
$$=\tan^{-1}\left(\dfrac{(r+1)-r}{1+(r+1)r}\right)=\tan^{-1}(r+1)-\tan^{-1}r$$
G.E$$=\displaystyle\sum^{\infty}_{r=0}\tan^{-1}(r+1)-\tan^{-1}(r)$$
$$=\tan^{-1}1-\tan^{-1}0+\tan^{-1}2-\tan^{-1}1+...…+\tan^{-1}\infty$$
$$=\tan^{-1}\infty -\tan^{-1}0$$
$$=\dfrac{\pi}{2}-0$$
$$=\dfrac{\pi}{2}$$.
The value of $${ sin }^{ -1 }(sin12)+{ cos }^{ -1 }(cos12)$$ is equal to :
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$$Zero$$
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$$24-2\pi $$
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$$4\pi -24$$
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None of these
Explanation
Given,
$$\sin ^{-1}(\sin 12)+\cos ^{-1}(\cos 12)$$
The principle values,
$$\sin ^{-1}x\in \left [ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right ],\forall x\in [-1,1]$$
$$\cos ^{-1}x\in \left [0,\pi \right ],\forall x\in [-1,1]$$
$$\therefore \sin ^{-1}(\sin 12)\neq 12\notin \left [ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right ]$$
$$\therefore \cos ^{-1}(\cos 12)\neq 12\notin \left [ 0,\pi \right ]$$
$$\therefore \sin ^{-1}(\sin 12)+\cos ^{-1}(\cos 12)$$
$$=\sin ^{-1}(\sin (12-4\pi ))+\cos ^{-1}(\cos (4\pi -12))$$
$$=(12-4\pi ) + (4\pi -12)$$
$$=0$$
$$\therefore \sin ^{-1}(\sin 12)+\cos ^{-1}(\cos 12)=0$$
The simplified form of $${ cos }^{ -1 }\left( \frac { 3 }{ 5 } { cos }x+\frac { 4 }{ 5 } { sin }x \right) $$ is :
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$${ tan }^{ -1 }\frac { 4 }{ 3 } -x$$
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$${ tan }^{ -1 }\frac { 1 }{ 3 } -x$$
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$${ tan }^{ -1 }\frac { 4 }{ 3 } +x$$
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None of these
$$\displaystyle \sum^{\infty}_{r=1}\tan^{-1}\left(\dfrac {3}{r^{2}-r+9}\right)$$ is -
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$$\dfrac {\pi}{3}$$
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$$\dfrac {\pi}{6}$$
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$$\dfrac {\pi}{2}$$
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$$\dfrac {\pi}{12}$$
$$If\,\cos \left( {2{{\sin }^{ - 1}}x} \right) = \frac{1}{9},\,then\,\,x\,\,is\,\,equal\,\,to$$
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$$Only\,\frac{2}{3}\,$$
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$$Only - \frac{2}{{3\,}}$$
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$$\,\frac{2}{3}, - \frac{2}{3}$$
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$$\frac{1}{3}$$
If $${ cos }^{ -1 }\dfrac { 3 }{ 5 } -{ sin }^{ -1 }\dfrac { 4 }{ 5 } ={ cos }^{ -1 }x$$, then x is equal to -
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0
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1
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-1
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None of these
Explanation
Given,
$$\cos ^{-1}\dfrac{3}{5}+\sin^{-1}\dfrac{4}{5}=\cos ^{-1}x$$
apply $$\cos$$ on both sides, we get,
$$\cos \left ( \cos ^{-1}\dfrac{3}{5}+\sin^{-1}\dfrac{4}{5} \right )=\cos (\cos ^{-1}x)$$
$$\mathrm{Use\:the\:following\:identity}:\quad \cos \left(s+t\right)=\cos \left(s\right)\cos \left(t\right)-\sin \left(s\right)\sin \left(t\right)$$
$$\cos \left(\cos ^{-1} \left(\dfrac{3}{5}\right)\right)\cos \left(\sin^{-1} \left(\dfrac{4}{5}\right)\right)-\sin \left(\cos ^{-1} \left(\dfrac{3}{5}\right)\right)\sin \left(\sin^{-1} \left(\dfrac{4}{5}\right)\right)=x$$
$$\dfrac{3}{5}\left ( \sqrt{1-\left (\dfrac{4}{5} \right )^2} \right )-\dfrac{4}{5}\left ( \sqrt{1-\left (\dfrac{3}{5} \right )^2} \right )=x$$
$$\dfrac{3}{5}\times \dfrac{3}{5}-\dfrac{4}{5}\times \dfrac{4}{5}=x$$
$$\dfrac{9}{25}-\dfrac{16}{25}=x$$
$$\therefore x=-\dfrac{7}{25}$$
Number of solution of the equation $$\tan^{-1}\left(\dfrac {1}{x-1}+\dfrac {1}{x-2}+\dfrac {1}{x-3}+\dfrac {1}{x-4} \right)+\cos^{-1}(x)=\dfrac {3\pi}{4}-\sin^{-1}(x)$$ are
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$$0$$
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$$1$$
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$$2$$
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$$3$$
The range of the function $$f(x)=\ell n\ (\sin^{-1}(x^{2}+x))$$ is
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$$\left[-\ell n \left(\sin^{-1}\dfrac {1}{4}\right),\ell n \dfrac {\pi}{4}\right]$$
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$$\left[-\ell n \dfrac {\pi}{4},\ell n \dfrac {\pi}{2}\right]$$
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$$\left(0,\ell n \dfrac {\pi}{2}\right]$$
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$$\left(-\infty ,\ell n \dfrac {\pi}{2}\right]$$
The trigonometric equation $$\sin^{-1}x=2\sin^{-1}a$$, has a solution for-
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$$\left|a\right|\le\dfrac{1}{\sqrt{2}}$$
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$$\dfrac{1}{2}<\left|a\right|<\dfrac{1}{\sqrt{2}}$$
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$$all\ real\ values\ of\ a$$
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$$\left|a\right|<\dfrac{1}{2}$$
Explanation
The trigonometric equation $$\text{sin}^{-1} x=2\text{sin}^{-1} a$$ will have a solution if $$2\text{sin}^{-1} a\in \bigg[-\dfrac{\pi}{2},\dfrac{\pi}{2}\bigg]$$ $$(\because\text{sin}^{-1} x\in [-\frac{\pi}{2},\frac{\pi}{2}])$$
$$\implies \text{sin}^{-1} a\in \bigg[-\dfrac{\pi}{4},\dfrac{\pi}{4}\bigg]$$
$$\implies a\in \bigg[-\sin \dfrac{\pi}{4},\sin \dfrac{\pi}{4}\bigg]$$
$$\implies a\in \bigg[-\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}\bigg]$$
$$\implies |a|\le \dfrac{1}{\sqrt{2}}$$
If $$f(x)=\cos^{-1}\left(\dfrac {\sqrt {2x^{2}+1}}{x^{2}+1}\right)$$, then range of $$f(x)$$ is
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$$[0,\pi]$$
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$$\left(0,\dfrac {\pi}{4}\right]$$
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$$\left(0,\dfrac {\pi}{3}\right]$$
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$$\left[0,\dfrac {\pi}{2}\right)$$
Explanation
Given,
$$f\left(x\right)=\cos ^{-1}\left(\dfrac{\sqrt{2x^2+1}}{x^2+1}\right)$$
Range of $$\dfrac{\sqrt{2x^2+1}}{x^2+1}$$
$$0< f(x)\leq1$$
$$\cos^{-1}(1)\leq \cos ^{-1}\left(\dfrac{\sqrt{2x^2+1}}{x^2+1}\right)<\cos^{-1}(0)$$
$$0\leq \cos ^{-1}\left(\dfrac{\sqrt{2x^2+1}}{x^2+1}\right)< \dfrac{\pi}{2}$$
Therefore the range is,
$$0\leq f(x)< \dfrac{\pi}{2}$$
Number of solution(s) to the equation$$ $$ $${\cos ^{ - 1}}x + {\sin ^{ - 1}}\left( {\dfrac{x}{2}} \right) = \dfrac{\pi }{6}\,$$ is/are
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$$0$$
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$$1$$
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$$2$$
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$$3$$
Explanation
Hence,
option $$(B)$$ is correct answer.
$$cos^{-1}\left(\dfrac{\pi}{3}+sec^{-1}(-2)\right)$$=
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-1
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1
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0
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None of these
Explanation
As we know that
$$\text{sec}^{-1}(-2)=\dfrac{2\pi}{3}$$
$$\therefore \cos \bigg(\dfrac{\pi}{3}+\text{sec}^{-1}(-2)\bigg)=\cos \bigg(\dfrac{\pi}{3}+\dfrac{2\pi}{3}\bigg)=\cos \pi=-1$$
The value of $$\cot\left(cosec^{-1}\dfrac{5}{3}+\tan^{-1}\dfrac{2}{3}\right)$$ is equal to-
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$$\dfrac{6}{17}$$
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$$\dfrac{3}{17}$$
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$$\dfrac{4}{17}$$
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$$\dfrac{5}{17}$$
$$\int _{ 0 }^{ \pi }{ \left[ cotx \right] dx,where\left[ \cdot \right] } $$ denotes the greatest integer function, is equal to:
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1
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-1
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$$-\dfrac { \pi }{ 2 } $$
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$$\dfrac { \pi }{ 2 }$$
$${ sec\quad h }^{ -1 }\left( sin\quad \theta \right) =$$
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$$log\left( tan\frac { \theta }{ 2 } \right) $$
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$$log\left( sin\frac { \theta }{ 2 } \right) $$
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$$log\left( cos\frac { \theta }{ 2 } \right)$$
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$$log\left( cot\frac { \theta }{ 2 } \right) $$
$$\cos^{ -1 }\left[\cos\left( -\frac { 17 }{ 15 } \pi \right) \right] $$ is equal to
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$$-\frac { 17\pi }{ 15 } $$
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$$\frac { 17\pi }{ 15 } $$
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$$\frac { 13\pi }{ 15 } $$
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$$\frac { -2\pi }{ 15 } $$
Find $$\displaystyle \int x.\sin xdx$$
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$$\sin x+x\cos x=C$$
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$$\sin x-x\cos x+C$$
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$$\sin x+x\sin x+C$$
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$$\sin x-x\sin x+C$$
If $$A=\tan^{1-}\left(\dfrac {x\sqrt {3}}{2k-x}\right)$$ and $$B=\tan^{-1}\left(\dfrac {2x-k}{k\sqrt {3}}\right)$$ then $$A.B=$$
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0%
$$0^{o}$$
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$$\pi /6$$
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$$\pi /4$$
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$$\pi /3$$
if x$$>0\quad then\quad { tanh }^{ -1 }\left( \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } \right)$$
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$${ log }_{ e }\left( 2x \right)$$
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$${ log }_{ e }x$$
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$${ log }_{ e }\left( 3x \right)$$
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$${ log }_{ e }\left( 5x \right)$$
$$if\quad x>0\quad then\quad { tanh }^{ -1 }\left( \frac { { x }^{ 2 }-1 }{ { x }^{ 2 }+1 } \right) $$
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$${ log }_{ e }\left( 2x \right) $$
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$${ log }_{ e }x$$
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$${ log }_{ e }\left( 3x \right) $$
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$${ log }_{ e }\left( 5x \right) $$
If $$\cos^{-1}x-\cos^{-1}(\dfrac {y}{2})=\alpha$$ $$ax^{2}-4xy\cos \alpha +y^{2}=$$
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$$-4\sin^{2}\alpha$$
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$$4\sin^{2}\alpha$$
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$$4$$
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$$2\sin 2\alpha$$
If $$\cot^{-1}{x}+\tan^{-1}\left (\dfrac{1}{3}\right)=\dfrac{\pi}{2}$$, then $$x$$ will be
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$$1$$
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$$3$$
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$$\dfrac {1}{3}$$
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None of these
Explanation
Given $$\text{cot}^{-1} x+\text{tan}^{-1}\bigg(\dfrac{1}{3}\bigg)=\dfrac{\pi}{2}$$
$$\implies \text{cot}^{-1} x=\dfrac{\pi}{2}-\text{tan}^{-1} \bigg(\dfrac{1}{3}\bigg)$$
$$\implies \text{cot}^{-1} x=\text{cot}^{-1}\bigg(\dfrac{1}{3}\bigg)$$
$$\implies x=\dfrac{1}{3}$$
If $$x={ \sin }^{ -1 }(\sin10) $$ and $$y={ \cos }^{ -1 }(\cos10)$$, then find $$y - x$$.
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$$\pi $$
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$$7\pi $$
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$$0$$
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$$10$$
Explanation
Given, $$x=sin^{-1}(sin 10)$$
$$x=3\pi -10$$
Also,
$$y=cos^{-1}(cos 10)$$
$$y=4\pi - 10$$
Therefore,
$$y-x=4\pi-10-3\pi+10 = \pi$$
State true or false.
$$\sin^{-1}x+\cos^{-1}x=\dfrac {\pi}{2}$$
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0%
True
0%
False
Explanation
As we know that
For $$y\in [-1,1],\text{sin}^{-1} y+\text{cos}^{-1} y=\dfrac{\pi}{2}$$
So $$\text{sin}^{-1} x+\text{cos}^{-1} x=\dfrac{\pi}{2}$$
So the relation is $$\text{True}$$
$${ \sin }^{ -1 }\left (\dfrac { 3 }{ 5 }\right )+{ \cos }^{ -1 }\left (\dfrac { 12 }{ 13 }\right )={ \sin }^{ -1 }\left (\dfrac { 56 }{ 65 } \right)$$
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0%
True
0%
False
$$\cos ^{ -1 }{ \left\{ \dfrac { 1 }{ 2 } { x }^{ 2 }+\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-\dfrac { { x }^{ 2 } }{ 4 } } \right\} =\cos ^{ -1 }{ \dfrac { x }{ 2 } -\cos ^{ -1 }{ x } } }$$ holds for:
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$$|x|\le 1$$
0%
$$x\in R$$
0%
$$0\le x\le 1$$
0%
$$-1\le x\le 0$$
Explanation
$$clearly, x/2\epsilon[-1,1]$$&$$x\epsilon[-1,1]$$
$$ x\epsilon [-2,2]$$&$$ x\epsilon[-1,1]$$
Intersection of above sells is [-1,1]
Thus $$,[\left | x \right |\leq 1]$$
$$\frac{\cos ^{-1}(41 / 49)}{\sin ^{-1}(2 / 7)}=$$
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4
0%
3
0%
2
0%
1
Explanation
$$\textbf{Step-1: Rewrite the Numerator observing Denominator}$$
$$\textbf{& apply multiple - angles of trigonometric function.}$$
$$\text{We have,}$$
$$\dfrac{\cos ^{-1}(41 / 49)}{\sin ^{-1}(2 / 7)}\\$$ $$\text{.....(i)}$$
$$\text{Now,}\\$$
$$\cos^{-1} (1- 2 (\dfrac{2}{7})^2)$$
$$\text{Put}$$ $$\sin \theta = \dfrac{2}{7}$$
$$\therefore$$ $$\cos^{-1}(1- 2 \sin^2 \theta)$$
$$\cos^{-1} (\cos 2\theta)$$ $$\boldsymbol{[\cos 2 \theta = 1 - 2 \sin^{2} \theta]}$$
$$ = 2\theta$$
$$\textbf{Step-2: Use the above results & get the required unknown.}$$
$$\dfrac{\cos ^{-1}(41 / 49)}{\sin ^{-1}(2 / 7)}\\$$ $$= \dfrac{2 \theta}{\theta} = 2$$ $$[\textbf{using (i)}]$$
$$\textbf{Hence, option - C is the answer}$$
The value of $$tan(\frac { 1 }{ 2 } { cos }^{ -1 }(\frac { \sqrt { 5 } }{ 3 } ))$$ is
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$$\frac { 3+\sqrt { 5 } }{ 2 } $$
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$$3-\sqrt { 5 } $$
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$$\frac { 1 }{ 2 } (3-\sqrt { 5 } )$$
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none of these
Explanation
Given,
$$y=\tan \left [ \dfrac{1}{2}\cos ^{-1}\left ( \dfrac{\sqrt{5}}{3} \right ) \right ]$$
Let, $$x=\cos ^{-1}\left ( \dfrac{\sqrt{5}}{3} \right )$$
$$\Rightarrow \cos x=\dfrac{\sqrt{5}}{3} $$
$$\therefore y=\tan \dfrac{1}{2}x$$
$$y=\tan \dfrac{x}{2}$$
$$y=\sqrt{\dfrac{1-\cos x}{1+\cos x}}$$
$$=\sqrt{\dfrac{1-\frac{\sqrt{5}}{3}}{1+\frac{\sqrt{5}}{3}}}$$
$$=\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}$$
rationalizing the factor, we get,
$$y=\dfrac{3-\sqrt{5}}{2}$$
$$\sin^{-1}\left(\dfrac{4}{5}\right)+\sin^{-1}\left(\dfrac{7}{25}\right)=\sin^{-1}\left(\dfrac{117}{125}\right)$$
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0%
True
0%
False
If tan (x + y) = 33 and x = $${ tan }^{ -1 }3$$, then y will be
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0.3
0%
$${ tan }^{ -1 }(1.3)$$
0%
$${ tan }^{ -1 }(0.3)$$
0%
$${ tan }^{ -1 }(\frac { 1 }{ 18 } )$$
Explanation
Given,
$$\tan (x+y)=33$$
$$x=\tan ^{-1}3$$
$$\Rightarrow \tan x=3$$
Formula,
$$\tan (x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$$
$$33=\dfrac{3+\tan y}{1-(3)\tan y}$$
$$33(1-3\tan y)=3+\tan y$$
$$33-99\tan y=3+\tan y$$
$$30=100\tan y$$
$$\tan y=\dfrac{30}{100}=0.3$$
$$\therefore y=\tan ^{-1}0.3$$
If $$sin^{-1}(\dfrac{1}{3}) + sin^{-1}(\dfrac{2}{3}) = sin^{-1} x$$, then x is equal to
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0%
$$0$$
0%
$$\dfrac{\sqrt{5}+4\sqrt{2}}{9}$$
0%
$$\dfrac{5\sqrt{2}-4\sqrt{5}}{9}$$
0%
$$\dfrac{\pi}{2}$$
Explanation
$$\sin^{-1}\dfrac{1}{3}+\sin^{-1}\dfrac{2}{3}\\\sin ^{-1}\left\{\dfrac{1}{3}\sqrt{1-\left(\dfrac{2}{3}\right)^2}+\dfrac{2}{3}\sqrt{1-\left(\dfrac{1}{3}\right)^2}\right\}\\\sin^{-1}\left\{\dfrac{1}{3}\left(\dfrac{\sqrt5}{3}\right)+\dfrac{2}{3}\left(\dfrac{\sqrt8}{3}\right)\right\}\\\sin^{-1}\left\{\dfrac{\sqrt5+4\sqrt2}{9}\right\}\\x=\dfrac{\sqrt5+4\sqrt2}{9}$$
The value of $$\sin ^{-1}(\sin 5\frac {\pi}{3})=$$
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$$-\frac {\pi}3$$
0%
$$\frac {\pi}3$$
0%
$$\frac {4\pi}3$$
0%
$$\frac {3\pi}3$$
Explanation
Given,
$$\sin ^{-1}\left (\sin 5\dfrac{\pi }{3} \right )$$
Let $$\theta =\sin ^{-1}\left (\sin 5\dfrac{\pi }{3} \right )$$
$$\sin \theta =\left (\sin 5\dfrac{\pi }{3} \right )=\sin \left ( 2\pi - \dfrac{\pi }{3}\right )$$
$$\Rightarrow \sin \theta =-\sin \dfrac{\pi }{3}$$
$$\therefore \theta =-\dfrac{\pi }{3}$$
The value of $$\displaystyle sec\left [ sin^{-1}\left (sin \dfrac{50\pi }{9} \right ) + cos^{-1}cos\left ( \dfrac{31\pi }{9} \right ) \right ]$$ is equal to
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0%
$$sec\dfrac{10\pi}{9}$$
0%
$$sec \ 9{\pi}$$
0%
$$-1$$
0%
$$1$$
Explanation
Given,
$$\sec \left [ \sin ^{-1}\left ( \sin \dfrac {50\pi }{9} \right ) +\cos ^{-1}\left ( \cos \dfrac {31\pi }{9} \right ) \right ]$$
$$=\sec \left [ \dfrac {50\pi }{9} + \dfrac {31\pi }{9}\right ]$$
$$=\sec \dfrac {81\pi }{9}$$
$$=\sec 9\pi$$
$$\tan { ^{ -1 }\left( 3/5 \right) } +\tan { ^{ -1 }\left( 1/4 \right) } =$$
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0%
$$0$$
0%
$$\pi /4$$
0%
$$3n/4$$
0%
None of these
Explanation
given, $${\tan}^{-1}{\dfrac{3}{5}}+{\tan}^{-1}{\dfrac{1}{4}}$$
$$={\tan}^{-1}{\left(\dfrac{\frac{3}{5}+\frac{1}{4}}{1-\frac{3}{5}\times\frac{1}{4}}\right)}$$
$$={\tan}^{-1}{\left(\dfrac{\frac{12+5}{20}}{1-\frac{3}{20}}\right)}$$
$$={\tan}^{-1}{\left(\dfrac{\frac{17}{20}}{\frac{20-3}{20}}\right)}$$
$$={\tan}^{-1}{\left(\dfrac{\frac{17}{20}}{\frac{17}{20}}\right)}$$
$$={\tan}^{-1}{\left(1\right)}$$
$$=\dfrac{\pi}{4}$$
$$\tan { ^{ -1 }\left( \tan { 3\pi /4 } \right) = } $$
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$$5\pi /4$$
0%
$$\pi /4$$
0%
$$-\pi /4$$
0%
None of these
Explanation
Let $$y={\tan}^{-1}{\left(\tan{\dfrac{3\pi}{4}}\right)}$$
$$\Rightarrow \tan{y}=\tan{\left(\pi-\dfrac{\pi}{4}\right)}$$
$$\Rightarrow \tan{y}=\tan{\dfrac{\pi}{4}}$$
$$\therefore y=\dfrac{\pi}{4},\pi+\dfrac{\pi}{4}$$
$$\Rightarrow y=\dfrac{\pi}{4},\dfrac{4\pi+\pi}{4}$$
$$\therefore y=\dfrac{\pi}{4},\dfrac{5\pi}{4}$$
The value of $$\sin^{-1}(\cos (\log_{2}(4\alpha -44)))$$ is
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$$\dfrac {\pi}{3}$$
0%
$$\dfrac {\pi}{2}$$
0%
$$0$$
0%
$$1$$
$$\tan (2\cos ^{-1}\frac 35)=$$_____
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$$\frac 83$$
0%
$$\frac {24}{25}$$
0%
$$\frac 7{25}$$
0%
$$\frac {-24}7$$
Explanation
Given,
$$\tan \left(2\cos ^{-1}\left(\frac{3}{5}\right)\right)$$
as $$\tan \left(2x\right)=\dfrac{2\tan \left(x\right)}{1-\tan ^2\left(x\right)}$$
$$=\dfrac{2\tan \left(\cos ^{-1} \left(\frac{3}{5}\right)\right)}{1-\tan ^2\left(\cos ^{-1} \left(\frac{3}{5}\right)\right)}$$
as $$\tan \left(\cos ^{-1}\left(x\right)\right)=\frac{\sqrt{1-x^2}}{x}$$
$$=\dfrac{2\left(\frac{\sqrt{1-\left(\frac{3}{5}\right)^2}}{\left(\frac{3}{5}\right)}\right)}{1-\left(\frac{\sqrt{1-\left(\frac{3}{5}\right)^2}}{\left(\frac{3}{5}\right)}\right)^2}$$
$$=\dfrac{2\cdot \frac{5\sqrt{-\left(\frac{3}{5}\right)^2+1}}{3}}{1-\left(\frac{5\sqrt{-\left(\frac{3}{5}\right)^2+1}}{3}\right)^2}$$
$$=\dfrac{2\cdot \frac{5\sqrt{-\left(\frac{3}{5}\right)^2+1}}{3}}{1-\frac{4^2}{3^2}}$$
$$=\dfrac{\frac{8}{3}}{1-\frac{4^2}{3^2}}$$
$$=\dfrac{8}{3\left(1-\frac{16}{9}\right)}$$
$$=-\dfrac{8}{\frac{7}{3}}$$
$$=-\dfrac{24}{7}$$
$${ \tan }^{ -1 }\left( \dfrac { 1 }{ 7 } \right) +{ \tan }^{ -1 }\left( \dfrac { 1 }{ 13 } \right) =\cos ^{ -1 }{ \left( \dfrac { 9 }{ 2 } \right) } $$
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0%
True
0%
False
The solution set of the equation$$2 cos^{ -1 } x = cot^{ -1 } \left(\dfrac { 2x^{ 2 } - 1 }{ 2x \sqrt { 1 x^{ 2 } } }\right) $$ is
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0%
$$\left(0, 1\right)$$
0%
$$\left(-1, 1\right)-{ 0 }$$
0%
$$\left(-1, 0\right)$$
0%
$$\left[-1, 1\right]$$
$${ \cos }^{ -1 }\left[ \cos \left( 2{ \cot }^{ -1 }\left( \sqrt { 2 } -1 \right) \right) \right]$$ is equal to
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0%
$$\sqrt{2}-1$$
0%
$$1-\sqrt{2}$$
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$$\pi/4$$
0%
$$3\pi/4$$
The number of solutions of the equation $$3\cos^{-1}x-\pi x-\dfrac {\pi}{2}=0$$
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0%
$$0$$
0%
$$1$$
0%
$$2$$
0%
$$infinite$$
If $$\alpha =\cos^{-1}\left(\dfrac{3}{5}\right),\beta=\tan^{-1}\left(\dfrac{1}{3}\right)$$ where $$0<\alpha,beta <\dfrac{\pi}{2},$$then $$\alpha -\beta $$ is equal to :
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0%
$$\sin^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)$$
0%
$$\tan^{-1}\left(\dfrac{9}{14}\right)$$
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$$\cos^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)$$
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$$\tan^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)$$
Explanation
$$\cos \alpha=\dfrac{3}{5},\tan \beta =\dfrac{1}{5}$$
$$\Rightarrow \tan \alpha =\dfrac{4}{3}$$
$$\Rightarrow tan (\alpha-\beta)=\dfrac{\dfrac{4}{3}-\dfrac{1}{3}}{1+\dfrac{4}{3}.\dfrac{1}{3}}=\dfrac{9}{3}$$
$$\Rightarrow \sin(\alpha-\beta)=\dfrac{9}{5\sqrt{10}}$$
$$\Rightarrow \alpha-\beta =\sin^{-1}\left(\dfrac{9}{5\sqrt{10}}\right)$$
If $$\;\sin {\;^{ - 1}}\dfrac{1}{x} = 2\;{\tan ^{ - 1}}\dfrac{1}{7} + {\cos ^{ - 1}}\dfrac{3}{5}$$, then $$x =$$ ___
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0%
$$\dfrac {24}{117}$$
0%
$$\dfrac {7}{3}$$
0%
$$\dfrac {125}{117}$$
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None of these
Explanation
$$2\tan ^{-1}\left(\dfrac{1}{7}\right)+\cos ^{-1}\left(\dfrac{3}{5}\right)$$
$$=\tan ^{-1}\left(\dfrac{1}{7}\right)+\tan ^{-1}\left(\dfrac{1}{7}\right)+\cos ^{-1}\left(\dfrac{3}{5}\right)$$
$$=\tan ^{-1}\left(\dfrac{\dfrac{1}{7}+\dfrac{1}{7}}{1-\dfrac{1}{7}\times \dfrac{1}{7}}\right)+\cos ^{-1}\left(\dfrac{3}{5}\right)$$ $$\left[\because \tan ^{-1}+\tan ^{-1}y \tan ^{-1x}=\tan ^{-1}\left(\dfrac{x+y}{1-xy}\right)\right]$$
$$=\tan ^{-1}\left(\dfrac{\dfrac{2}{7}}{1-\dfrac{1}{49}}\right)+\theta$$
$$=\tan ^{-1}\left(\dfrac{7}{24}\right)+\tan ^{-1}\left(\dfrac{4}{3}\right)$$
$$=\tan ^{-1}\left(\dfrac{\dfrac{7}{24}+\dfrac{4}{3}}{1-\dfrac{7}{24}\times \dfrac{4}{3}}\right)$$
$$=\tan ^{-1}\left(\dfrac{117}{44}\right)$$
$$\because \sin ^{-1}\left(\dfrac{1}{x}\right)=\tan ^{-1}\left(\dfrac{117}{44}\right)$$
Let $$\theta =\tan ^{-1}\left(\dfrac{117}{44}\right)$$
$$\therefore \tan \theta=\dfrac{117}{44}$$
$$\therefore \sin \theta=\dfrac{117}{\sqrt{117^2+44^2}}=\dfrac{117}{125}$$
$$\therefore \theta =\sin ^{-1}\left(\dfrac{117}{125}\right)$$
$$\therefore \sin ^{-1}\left(\dfrac{1}{x}\right)=\sin ^{-1}\left(\dfrac{117}{125}\right)$$
On comparing
$$\dfrac{1}{x}=\dfrac{117}{125}$$
$$\boxed{\therefore x=\dfrac{125}{117}}$$
If $$\cos ^{ -1 }{ x } -\cos ^{ -1 }{ \cfrac { y }{ 2 } } =\alpha $$ where $$-1-1\le x\le 1,-2\le y\le 2,x\le \cfrac { y }{ 2 } $$ then for all $$4{ x }^{ 2 }-4xy\cos { \alpha } +{ y }^{ 2 }$$ is equal to
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$$4\sin ^{ 2 }{ \alpha } -2{ x }^{ 2 }{ y }^{ 2 }$$
0%
$$4\cos ^{ 2 }{ \alpha } +2{ x }^{ 2 }{ y }^{ 2 }$$
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$$4\sin ^{ 2 }{ \alpha } $$
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$$2\sin ^{ 2 }{ \alpha } $$
Explanation
$$\cos { \left( \cos ^{ -1 }{ x } -\cos ^{ -1 }{ \cfrac { y }{ 2 } } \right) } =\cos { \alpha } $$
$$\cos { \alpha } \Rightarrow x\times \cfrac { y }{ 2 } +\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-\cfrac { { y }^{ 2 } }{ 4 } } \Rightarrow { \left( \cos { \alpha } -\cfrac { xy }{ 2 } \right) }^{ 2 }=\left( 1-{ x }^{ 2 } \right) \left( 1-\cfrac { { y }^{ 2 } }{ 4 } \right) $$
$${ x }^{ 2 }+\cfrac { { y }^{ 2 } }{ 4 } -xy\cos { \alpha } =1-\cos ^{ 2 }{ \alpha } =\sin ^{ 2 }{ \alpha } \quad $$
The value of $$tan \left (\cos^{-1} \dfrac {3}{5} + \tan^{-1} \dfrac {1}{4}\right )$$ is
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0%
$$\dfrac {19}{8}$$
0%
$$\dfrac {8}{19}$$
0%
$$\dfrac {19}{12}$$
0%
$$\dfrac {3}{4}$$
Explanation
Given, $$\tan{\left({\cos}^{-1}{\dfrac{3}{5}}+{\tan}^{-1}{\dfrac{1}{4}}\right)}$$
Let, $${\cos}^{-1}{\dfrac{3}{5}}=a$$
$$\Rightarrow\,\cos{a}=\dfrac{3}{5}$$
$$\Rightarrow\,{\sin}^{2}{a}=1-{\cos}^{2}{a}=1-\dfrac{9}{25}=\dfrac{25-9}{25}=\dfrac{16}{25}$$
$$\therefore\,\sin{a}=\dfrac{4}{5}$$
$$\Rightarrow\,\tan{a}=\dfrac{\sin{a}}{\cos{a}}=\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}=\dfrac{4}{3}$$
$$\Rightarrow\,a={\tan}^{-1}{\dfrac{4}{3}}$$
$$\therefore\,\tan{\left({\cos}^{-1}{\dfrac{3}{5}}+{\tan}^{-1}{\dfrac{1}{4}}\right)}$$
$$=\tan{\left(a+{\tan}^{-1}{\dfrac{1}{4}}\right)}$$ where $$a={\cos}^{-1}{\dfrac{3}{5}}$$
$$=\tan{\left({\tan}^{-1}{\dfrac{4}{3}}+{\tan}^{-1}{\dfrac{1}{4}}\right)}$$ where $$a={\cos}^{-1}{\dfrac{3}{5}}={\tan}^{-1}{\dfrac{4}{3}}$$
We know that $${\tan}^{-1}{A}+{\tan}^{-1}{B}={\tan}^{-1}{\left(\dfrac{A+B}{1-AB}\right)}$$
$$=\tan{\left({\tan}^{-1}{\left(\dfrac{\dfrac{4}{3}+\dfrac{1}{4}}{1-\dfrac{4}{3}\times\dfrac{1}{4}}\right)}\right)}$$
$$=\tan{\left({\tan}^{-1}{\left(\dfrac{\dfrac{16+3}{12}}{1-\dfrac{4}{12}}\right)}\right)}$$
$$=\tan{\left({\tan}^{-1}{\left(\dfrac{\dfrac{19}{12}}{\dfrac{12-4}{12}}\right)}\right)}$$
$$=\tan{\left({\tan}^{-1}{\dfrac{19}{8}}\right)}$$
$$=\dfrac{19}{8}$$
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Practice Class 12 Commerce Maths Quiz Questions and Answers
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