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Answers posted by thanvigandhi_1
Questions
3013
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Prove that \[cot^{-1} \left ( \frac {{\sqrt {1+sinx}}+{\sqrt {1-sinx}}}{{\sqrt {1+sinx}}-{\sqrt {1-sinx}}} \right ) = \frac {x}{2} , x \in \left ( 0, \frac{\pi}{4} \right )\]
answered
Feb 23, 2013
Toolbox: \(sin^2\large\frac{x}{2}+cos^2\large\frac{x}{2}=1\) \((a+b)^2=a^2+b^2+2ab\)...
0
votes
Prove that \[tan^{-1} \sqrt{x} = \frac{1}{2} cos^{-1} \;\; \bigg(\frac{1-x}{1+x}\bigg), x\in [0,1]\]
answered
Feb 23, 2013
Toolbox: \(\large\frac{1-tan^2\theta}{1+tan^2\theta}\)\(=cos2\theta\) Given $tan^{...
0
votes
Prove that \[tan^{-1} \frac{1}{5}+ tan^{-1} \frac{1}{7}+tan^{-1}\frac{1}{3}+tan^{-1} \frac{1}{8}= \frac{\pi}{4}\]
answered
Feb 23, 2013
Toolbox: \( tan^{-1}x=tan^{-1}y=tan^{-1} \bigg( \large\frac{x+y}{1-xy} \bigg) xy < 1\) ...
0
votes
What does $\sin^{-1} \frac{5}{13}+\cos^{-1} \frac {3}{5}$ reduce to?
answered
Feb 23, 2013
Toolbox:$\sin^{-1}x=\tan^{-1}\large \frac{x}{\sqrt{1-x^2}}$$\cos^{-1}x=\tan^{-1}\large \frac{\sqrt{1...
0
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Prove that \[ cos^{-1} \frac {12}{13} +sin^{-1} \frac{3}{5}=sin^{-1} \frac{56}{65}\]
answered
Feb 23, 2013
Toolbox: \( cos^{-1}x=sin^{-1}\sqrt{1-x^2}\) \( sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ ...
0
votes
Prove that \[cos^{-1} \frac {4}{5} +cos^{-1} \frac{12}{13}=cos^{-1} \frac{33}{65}\]
answered
Feb 23, 2013
Toolbox: \( cos^{-1}x+cos^{-1}y=cos^{-1} (xy- \sqrt{1-x^2} \sqrt{1-y^2} )\) Given ...
0
votes
Prove that \[sin^{-1} \frac{8}{17} +sin^{-1} \frac{3}{5} =tan^{-1} \frac{77}{36}\]
answered
Feb 23, 2013
Toolbox: \( sin^{-1}x=tan^{-1}\large\frac{x}{\sqrt{1-x^2}}\) \( tan^{-1}x+tan^{-1}y ...
0
votes
What does $ 2\sin^{-1} \frac{3}{5}$ reduce to
answered
Feb 23, 2013
Toolbox: \( sin^{-1}x=tan^{-1}\large\frac{x}{\sqrt{1-x^2}}\) \( 2tan^{-1}x=tan^{-1}\large\frac...
0
votes
Find the value of the following:
$ tan^{-1} \bigg(tan \frac{7\pi}{6}\bigg)$
answered
Feb 23, 2013
Toolbox:Principal interval of tan is $(\large-\frac{\pi}{2},\large\frac{\pi}{2})$$\tan(\pi+x)=\tan x...
0
votes
Find the value of: $\cos^{-1} (\cos \large\frac{13\pi}{6} ) $
answered
Feb 23, 2013
Toolbox:Principal interval of cos is $[0,\pi]$$\cos(2\pi+x)=\cos x$Since $\large\frac{13\pi}{6}$ is ...
0
votes
Find the value of $ \tan^{-1} \sqrt 3 - \cot^{-1} (-\sqrt 3)$
answered
Feb 23, 2013
Toolbox:$\tan^{-1}\sqrt{3}=\large\frac{\pi}{3}$$\cot^{-1}-\sqrt{3}=\pi-\large\frac{\pi}{6}$ Ans (B...
0
votes
Find the values of each of the expressions in $ \sin \bigg(\frac{\pi}{3} - \sin^{-1} ( -\frac{1}{2} ) \bigg) $ is equal to
answered
Feb 23, 2013
Toolbox:$\sin^{-1}(-\large\frac{1}{2})=-\large\frac{\pi}{6}$$\sin\large\frac{\pi}{2}=1$Given $\sin \...
0
votes
Find the value of the expression $\cos^{-1} \bigg(\cos \frac{\large 7\pi}{\large 6} \bigg) $
answered
Feb 23, 2013
Toolbox:$\cos(\pi+x)=-\cos x$$\cos(\pi+x)=cos(\pi-x)$Principal interval of cos is [0,\(\pi\)]Given $...
0
votes
Find the value of the expression in \[ tan^{-1} \bigg( tan \frac{3\pi}{4} \bigg) \]
answered
Feb 23, 2013
Toolbox:\(tan(\pi-x)=-tanx\)\(tan^{-1}(-1)=-\frac{\pi}{4}\)Given $tan^{-1} \bigg( tan \frac{3\pi}{4}...
0
votes
Find the values of the expression $ \tan \bigg( \sin^{-1} \frac{3}{5} + \cot^{-1} \frac{3}{2} \bigg) $
answered
Feb 23, 2013
Toolbox:$ \sin^{-1}x=\tan^{-1}\large \bigg(\frac{x}{\sqrt{1-x^2}}\bigg) $$\cot^{-1}x=\tan^{-1}\frac{...
0
votes
Find the values of \[ sin^{-1} \bigg( sin \frac{2\pi}{3} \bigg) \]
answered
Feb 23, 2013
Toolbox:$sin(\pi - x) = sinx$The range of the principal value of $\sin^{-1}x$ is $\left [ -\frac{\pi...
0
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If $ \tan^{-1} \frac{\large x-1}{\large x-2} + \tan^{-1} \frac{\large x+1}{\large x+2} = \frac{\large \pi}{\large 4} $ then find the value of $x$
answered
Feb 22, 2013
Toolbox:$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\large\frac{x+y}{1-xy}\: xy < 1$$\tan \large\frac{\pi}{4}...
0
votes
Find the value of $x$ if \( sin \bigg ( sin^{-1} \frac{1}{5} + cos^{-1} \: x \bigg) = 1\).
answered
Feb 22, 2013
Toolbox: \( sin^{-1}x+cos^{-1}x=\large\frac{\pi}{2}\) \( sin\large\frac{\pi}{2}=1\: ...
0
votes
Find the value of \[ tan\frac{1}{2} \bigg[ sin^{-1} \frac{2x}{1+x^2} + cos^{-1} \frac{1-y^2}{1+y^2} \bigg], | x | < 1, y > 0 \: and \: xy < 1\]
answered
Feb 22, 2013
Toolbox:$\large \frac{2\tan\theta}{1+\tan^2\theta}=\sin2\theta$$\large \frac{1-\tan^2\theta}{1+\tan^...
0
votes
Find the value of \[ \cot \;( \tan^{-1}a + \cot^{-1}a) \]
answered
Feb 22, 2013
Toolbox:\( tan^{-1}x+cot^{-1}x=\frac{\pi}{2}\) for all real xcot\(\frac{\pi}{2}=0\)Substituting, \( ...
0
votes
Find the value of $ \tan^{-1} \bigg[ 2\cos \bigg(2\sin^{-1}\frac{1}{2}\bigg) \bigg] $
answered
Feb 22, 2013
Toolbox:$ \sin\large\frac{\pi}{6}=\frac{1}{2}$$ \cos\large\frac{\pi}{3}=\frac{1}{2}$$\tan^{-1}1=\lar...
0
votes
Write the following function in the simplest form:\[ tan^{-1} \bigg( \frac{3a^2x-x^3}{a^3-3ax^2} \bigg), a > 0 ; \frac{-a}{\sqrt 3} \leq x \leq \frac{a}{\sqrt 3} \]
answered
Feb 22, 2013
Toolbox:\( tan3\theta= \Large \frac{3tan\theta-tan^3\theta}{1-3tan^2\theta}\)$tan^{-1}tan\; \theta =...
0
votes
Write the following in the simplest form: $ \tan^{-1} \large \frac{x}{\sqrt {a^2 - x^2}},$$ | x | < a $
answered
Feb 22, 2013
Toolbox: \( 1-sin^2\theta=cos^2\theta\)Given $tan^{-1} \large \frac{x}{\sqrt {a^2 - x^2}}$,$ | x ...
0
votes
Write the following function in the simplest form: \[ tan^{-1} \bigg( \frac{cos\: x - sin\: x}{cos\: x + sin\: x} \bigg) , 0 < x < \pi \]
answered
Feb 22, 2013
Toolbox:\( \large \frac{cosx-sinx}{cosx+sinx}\) \(=\large \frac{1-tanx}{1+tanx}\)\(tan(A-B)=\large \...
0
votes
Write the following function in the simplest form: \[ tan^{-1} \bigg( \sqrt {\frac{1-cos \: x}{1 + cos \: x}} \bigg), x < \pi\]
answered
Feb 22, 2013
Toolbox: \( 1-cos\theta = 2sin^2\frac{\theta}{2}\) \( 1+cos\theta = 2cos^2\frac{\the...
0
votes
Write the following function in the simplest form: \[ tan^{-1} \frac{1}{\sqrt{x^2 - 1}}, | x | > 1 \]
answered
Feb 22, 2013
Toolbox: \( sec^2\theta-1=tan^2\theta\) \( cot\theta = tan ( \large\frac{\pi}{2}-\th...
0
votes
Prove the following \[3\cos^{-1}x=\cos^{-1}(4x^3-3x), \;x\;\in\bigg[\frac{1}{2},1\bigg]\]
answered
Feb 22, 2013
Toolbox: \(cos3A=4cos^3A-3cosA\) \( cos^{-1}cosx=x \: \: \: if\: x \in [ 0\: \pi ] \...
0
votes
Prove the following\[3\sin^{-1}x=\sin^{-1}(3x-4x^3)\;x\in\bigg[-\frac{1}{2},\frac{1}{2}\bigg]\]
answered
Feb 22, 2013
Toolbox: \( sin\: 3A=3\: sinA-4sin^3A\) \( sin^{-1}sinx=x\: \: \: \: if\: x \: \in\:...
0
votes
Solve the following $\tan^{-1} \sqrt{3} -sec^{-1} (-2)$:
answered
Feb 22, 2013
Ans : B$\frac{\pi}{3} - \bigg( \pi-\frac{\pi}{3} \bigg) = \frac{-\pi}{3} $
0
votes
Find the value of the following:\[cos^{-1} \frac {1} {2} +2 sin^{-1}\frac {1} {2} \]
answered
Feb 22, 2013
Ans : \( \frac{\pi}{3}+2.\frac{\pi}{6}=\frac{2\pi}{3} \)
0
votes
Find the value of the following: \[ tan^{-1} (1) \;+ \; cos^{-1} \frac {-1} {2} + sin^{-1} \frac {-1} {2} \]
answered
Feb 22, 2013
Ans : \( \large\frac{\pi}{4}+\bigg( \pi-\large\frac{\pi}{3} \bigg)-\large\frac{\pi}{6} = \large...
0
votes
Find the principal values of the following:
$ \mathrm{cosec}^{-1} (-\sqrt 2) $
answered
Feb 22, 2013
Ans : $ cosec^{-1}cosec \bigg( \large\frac{-\pi}{4} \bigg) = \large\frac{-\pi}{4} $
0
votes
Find the principal value of the following: $cos^{-1} \bigg( -\frac {1} {\sqrt 2} \bigg) $
answered
Feb 22, 2013
Ans : $ cos^{-1} \bigg( cos \bigg( \pi-\frac{\pi}{4} \bigg) \bigg) = \frac{3\pi}{4} $
0
votes
Find the principal values of the following : $\cot^{-1} (\sqrt 3) $
answered
Feb 22, 2013
Toolbox:The range of the principle value of $\cot^{-1} x $ is $[0, \pi]$Let $\cot^{-1 } \sqrt 3 = x ...
0
votes
Find the principal values of the following: $sec^{-1} \bigg( \frac {2} {\sqrt 3} \bigg) $
answered
Feb 22, 2013
Ans : $ sec^{-1}sec\frac{\pi}{6}=\frac{\pi}{6} $
0
votes
Find the principal values of the following: $tan^{-1} (-1)$
answered
Feb 22, 2013
Ans : $( tan^{-1}tan \frac{-\pi}{4}=\frac{-\pi}{4} )$
0
votes
Find the principal values of the following: $cos^{-1} \bigg(-\frac {1} {2} \bigg ) $
answered
Feb 22, 2013
Ans : $( cos^{-1} \bigg( cos \bigg( \pi - \large\frac{\pi}{3} \bigg) \bigg)=\large\frac{2 \pi}{3})$
0
votes
Find the principal values of the following:
$ tan^{-1} (-\sqrt3)$
answered
Feb 22, 2013
Ans : $tan^{-1}tan \large\frac{-\pi}{3}=\large\frac{-\pi}{3} $
0
votes
Find the principal values of the following:
$cos^{-1} \bigg(\large\frac { \sqrt 3} {2}\bigg)$
answered
Feb 22, 2013
Ans : $ cos^{-1} \bigg( cos\large\frac{\pi}{6} \bigg) = \large\frac{\pi}{6} $
0
votes
Find the principal values of the following:
$ sin^{-1} \bigg( -\frac {1} {2}\bigg) $
answered
Feb 22, 2013
Toolbox:Principal interval of $( \sin x) is \bigg[ \large\frac{\pi}{2}, \large\frac{\pi}{2} \bigg] $...
0
votes
Find the value of x if $cosec^{-1}x=2\cot^{-1}7+\cot^{-1}\frac{3}{4}$
answered
Feb 20, 2013
Toolbox:$\cot^{-1}x=\tan^{-1}\large\frac{1}{x} $$ 2\tan^{-1}x=\tan^{-1}\large\frac{2x}{1-x^2} $$\tan...
0
votes
Find the value of $\cos\bigg\{\cos^{-1}\bigg(\frac{-\sqrt {3}}{2}\bigg)+\frac{\pi}{6}\bigg\}$
answered
Feb 20, 2013
Toolbox:$(cos(\pi-\theta)=-cos\theta)$Principal interval of cos is $( [0,\pi])$$(cos\large\frac{\pi}...
0
votes
Find the value of $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3.$
answered
Feb 20, 2013
Toolbox: \( tan^{-1}1=\large\frac{\pi}{4} \) \( tan^{-1}x+tan^{-1}y=tan^{-1}\large\f...
0
votes
Solve for $x$: $\sin\bigg[\sin^{-1}\frac{1}{3}+\cos^{-1}x\bigg]=1$
answered
Feb 20, 2013
Toolbox: \( sin^{-1}x+cos^{-1}x=\large\frac{\pi}{2}\) \( sin\large\frac{\pi}{2}=1\: ...
0
votes
Simplify $\tan^{-1}\big(\large \frac{\cos x}{1+\sin x}\big)$
answered
Feb 20, 2013
Toolbox:$\cos x=\cos^2\large\frac{x}{2}-\sin^2\large\frac{x}{2}$$\large\frac{1-\tan x}{1+\tan x}=\ta...
0
votes
Considering the principal solutions, find the number of solutions of $ \tan^{-1}2x+\tan^{-1}3x=\frac{\pi}{4}$
answered
Feb 20, 2013
Toolbox: \( tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}\:\:xy<1 \) \(tan^{...
0
votes
Find the principal value of $\sin^{-1}\bigg(\frac{-\sqrt {3}}{2}\bigg)+\cos^{-1}\bigg(\frac{-\sqrt {3}}{2}\bigg)$
answered
Feb 20, 2013
Toolbox: \(sin^{-1}x+cos^{-1}x=\large\frac{\pi}{2}\) for all x in the domain  ...
0
votes
If $\cos^{-1}x=\tan^{-1}x,\;show\;that\;\sin(\cos^{-1}x)=x^2$
answered
Feb 20, 2013
Toolbox: \( cos^{-1}x = tan^{-1} \large\frac{\sqrt{1-x^2}}{x} \) \( sincos^{-1}x=\sq...
0
votes
If $x>0$ and $\sin^{-1}\bigg(\large{\frac{5}{x}}\bigg)+\sin^{-1}\bigg(\frac{12}{x}\bigg)=\large\frac{\pi}{2},$ than find the value of $x$.
answered
Feb 20, 2013
Toolbox: \( sin^{-1}x=cos^{-1}\sqrt{1-x^2} \) \(cos\theta=sin(\large\frac{\pi}{2}-\t...
0
votes
If $x,y,z \in [-1,1]\;$ such that $\;\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\frac{3\pi}{2},$ find the value of $x^{2006}+y^{2007}+z^{2008}-\Large \frac{9}{x^{2006}+y^{2007}+z^{2008}}$
answered
Feb 20, 2013
Toolbox:Principal interval of $\sin^{-1}x \;is\;[-\large\frac{\pi}{2},\frac{\pi}{2}]$$\sin\large\fra...
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