Answers posted by thanvigandhi_1

0 votes

Find the value of $tan^{-1}\frac{-1}{\sqrt 3}+cot^{-1}\frac{1}{\sqrt 3}-tan^{-1}sin\frac{\pi}{2}$

answered Mar 2, 2013

Toolbox: $sin\large\frac{\pi}{2}=1$ $-tan\theta=tan(-\theta)$ \(\large\fra...

0 votes

Prove that \[cot(\frac{\pi}{4}-2cot^{-1}3)=7\]

answered Mar 2, 2013

Toolbox: $ 2cot^{-1}x= cot^{-1} \bigg( \large\frac{x^2-1}{2x} \bigg)$ \(cot\large\...

0 votes

Evaluate $cos\bigg(cos^{-1}\bigg(\frac{-\sqrt{3}}{2}\bigg)+\frac{\pi}{6}\bigg)$

answered Mar 2, 2013

Toolbox:$(-cos\theta=cos(\pi-\theta))$Principal interval of cos is $([0,\pi])$$(cos\large\frac{\pi}{...

0 votes

Prove that $tan^{-1}\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\frac{\pi}{4}+\frac{1}{2}cos^{-1}x^2$

answered Mar 2, 2013

Toolbox: $ put \: x^2 = cos\theta$ $ \Rightarrow \theta = cos^{-1}x^2$ \( ...

0 votes

Simplify$ 2\tan^{-1} \bigg(\large \frac{1}{5} \bigg) + \sec^{-1} \bigg( \large\frac{5\sqrt 2}{7} \bigg) + 2\: tan^{-1} \bigg( \frac{1}{8} \bigg) $

answered Mar 1, 2013

Toolbox: $ tan^{-1}x+tan^{-1}y=tan^{-1}\bigg(\large\frac{x+y}{1-xy} \bigg) xy<1$ ...

0 votes

Prove that $ tan^{-1} \bigg(\frac{\large 1}{\large 4} \bigg) + tan^{-1} \bigg( \frac{\large 2}{\large 9} \bigg ) = \frac{\large 1}{\large 2} cos^{-1} \bigg( \frac{3}{5} \bigg). $

answered Mar 1, 2013

Toolbox: $ tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}\:\:\:xy<1$ \( ta...

0 votes

Solve for $ x : \sin^{-1} (1-x)-2 \sin^{-1}\: x = \large\frac{\pi}{2} $

answered Mar 1, 2013

Toolbox: $ sin\bigg(\large \frac{\pi}{2}+y \bigg) = cosy$ $ cos2\theta = 1-2sin^2\theta$ ...

0 votes

Write the following functions in the simplest form : \[ (i) tan^{-1} \bigg( \frac{\sqrt{1+x^2} - 1}{x} \bigg), x \neq 0 \] \[(ii) tan^{-1} \frac{1}{\sqrt{x^2-1}} ,| x | > 1 \]

answered Mar 1, 2013

Toolbox: $ 1-cos\theta=2sin^2\large\frac{\theta}{2}$ \( sin\theta=2sin\large\frac{...

0 votes

Prove that : $ tan^{-1}\bigg(\frac{1}{3} \bigg)+tan^{-1}\bigg(\frac{1}{5}\bigg)+tan^{-1}\bigg(\frac{1}{7}\bigg)+tan^{-1}\bigg(\frac{1}{8}\bigg)=\large\frac{\pi}{4} $

answered Mar 1, 2013

Toolbox: $ tan^{-1}x=tan^{-1}y=tan^{-1} \bigg( \large\frac{x+y}{1-xy} \bigg) xy < 1$ Given...

0 votes

Prove that : $ \cos^{-1}\large \frac{4}{5}+\tan^{-1}\large\frac{3}{5}$$=\tan^{-1}\large\frac{27}{11} $

answered Mar 1, 2013

Toolbox: $ cos^{-1}x=tan^{-1}\large\frac{\sqrt{1-x^2}}{x}$ \( tan^{-1}x+tan^{-1}y=...

0 votes

If $ cos^{-1}x+cos^{-1}y+cos^{-1}z=\pi$ prove that $ x^2+y^2+z^2+2xyz=1$

answered Mar 1, 2013

Toolbox: $ cos^{-1}x+cos^{-1}y=cos^{-1} [ xy-\sqrt{1-x^2} \sqrt{1-y^2} ]$ \( cos^{...

0 votes

Solve $ 2tan^{-1}(cos \: x)=tan^{-1}(2 cosec \: x)$

answered Mar 1, 2013

Toolbox: $ 2tan^{-1}y=tan^{-1}\large\frac{2y}{1-y^2}\:\:|y|<1$ $1-cos^2x=sin^2x$ Give...

0 votes

Prove that : \[ sin^{-1}\frac{3}{5}+sin^{-1}\frac{8}{17}=cos^{-1}\frac{36}{85} \]

answered Mar 1, 2013

Toolbox: $ sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt {1-y^2}+y\sqrt{1-x^2} \bigg]$ ...

0 votes

If $ tan^{-1} \bigg(\large \frac{x-1}{x-2} \bigg) $$+ \tan^{-1} \bigg(\large \frac{x+1}{x+2} \bigg) =\large \frac{\pi}{4}, $ find the value of x.

answered Mar 1, 2013

Toolbox: $tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}\: xy < 1$ tan\(\large\frac{\...

0 votes

Solve for x : \[ tan^{-1} \bigg( \frac{2x}{1-x^2} \bigg) +cot^{-1} \bigg( \frac{1-x^2}{2x} \bigg) = \frac{\pi}{3}, x > 0\]

answered Mar 1, 2013

Toolbox: Put $ x = tan\theta \Rightarrow \theta=tan^{-1}x$ \( \large\frac{2tan\the...

0 votes

Prove that $ \tan^{-1} \large\frac{1}{7}$$ + 2\tan^{-1} \large\frac{1}{3} = \frac{\pi}{4} $

answered Mar 1, 2013

Toolbox: $ 2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2}$ \(tan^{-1}x+tan^{-1}y=tan^{-...

0 votes

Solve for x : $ \tan^{-1} \bigg(\large \frac{x-1}{x+1} \bigg) + \tan^{-1} \bigg( \frac{2x-1}{2x+1} \bigg) = \tan^{-1} \bigg( \frac{23}{36} \bigg). $

answered Mar 1, 2013

Toolbox: $ tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy} , xy < 1$ \( ta...

0 votes

Prove that $ 2\tan^{-1} \large\frac{1}{5}$$ + \tan^{-1} \large\frac{1}{8}$$ = \tan^{-1}\large \frac{4}{7} $

answered Mar 1, 2013

Toolbox: $ 2tan^{-1}x=tan^{-1}\bigg[ \large\frac{2x}{1-x^2}\bigg]$ \( tan^{-1}x+ta...

0 votes

Prove that : $\sin^{-1}\frac{3}{5} + \sin^{-1}\frac{8}{17}=\sin^{-1}\frac{77}{85} $

answered Mar 1, 2013

Toolbox:$ sin^{-1}x+sin^{-1}y$$= sin^{-1} \bigg[ x\sqrt{1-y^2}+y\sqrt{1-x^2} \bigg]$$L.H.S sin^{-1}...

0 votes

If $ \cos^{-1}\large\frac{x}{a}$$+\cos^{-1}\large\frac{y}{b}= \alpha $, prove that $\large \frac{x^2}{a^2}-\frac{2xy}{ab}$$\cos \alpha+\large\frac{y^2}{b^2}$$=\sin^2\alpha. $

answered Mar 1, 2013

Toolbox:$ cos^{-1}x+cos^{-1}y=cos^{-1} \bigg[ xy-\sqrt{1-x^2}\sqrt{1-y^2} \bigg] $\( cos^{-1}\frac...

0 votes

Show that $ \sin^{-1}\large\frac{12}{13}$$+\cos^{-1}\large\frac{4}{5}$$+\tan^{-1}\large\frac{63}{16} $$= \pi $

answered Feb 28, 2013

Toolbox: $ sin^{-1}\large\frac{12}{13}=tan^{-1}\large\frac{12}{5}$ \( cos^{-1}\lar...

0 votes

Solve for x : $ 2\: tan^{-1}(\cos x)=tan^{-1}(2 cosec x) $

answered Feb 28, 2013

Toolbox: $ 2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2}$ \( tan^{-1}\large\frac{2co...

0 votes

Solve : $ \tan^{-1}2x+tan^{-1}3x = \large\frac{\pi}{4}. $

answered Feb 28, 2013

Toolbox: $ tan^{-1}x+tan^{-1}y=tan^{-1} \bigg(\large \frac{x+y}{1-xy} \bigg)$ \(...

0 votes

Simplify : $ \cos^{-1} \bigg(\large \frac{3}{5}$$ \cos \theta +\large \frac{4}{5}$$ sin \theta \bigg) $

answered Feb 28, 2013

Toolbox: Take $ \large\frac{3}{5}=cosA$ $ \Rightarrow \large\frac{4}{5}=sinA$ ...

0 votes

Show that : $ \tan\large\frac{1}{2} \bigg[ \sin^{-1}\frac{2x}{1+x^2} + \cos^{-1}\large\frac{1-y^2}{1+y^2} \bigg] =\large \frac{x+y}{1-xy}$$, |x| < 1, y > 0 \: xy < 1 $

answered Feb 28, 2013

Toolbox: Put $ x= tan\alpha \: and \: y=tan \beta$ \(\large \frac{2tan\alpha}{1+ta...

0 votes

Solve for x : $ 2\tan^{-1}(\sin x)=\tan^{-1} (2\sec x),0 < x < \large\frac{\pi}{2} $

answered Feb 28, 2013

Toolbox:$ 2tan^{-1}x=tan^{-1}\frac{2x}{1-x^2}$By taking sinx in the place of x in the above formul...

0 votes

Solve $ \tan^{-1}(x+1)+tan^{-1}(x-1)=tan^{-1}\large\frac{8}{31} $.

answered Feb 28, 2013

Toolbox: $ tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}, xy< 1$By taking x+1 in the pl...

0 votes

Find the value of $ \tan{-1} \bigg(\large \frac{x}{y} \bigg)-\tan^{-1} \bigg(\large \frac{x-y}{x+y} \bigg). $

answered Feb 28, 2013

Toolbox: \( tan^{-1}\alpha-tan^{-1}\beta=tan^{-1}\large\frac{\alpha-\beta}{1+\alpha\beta}\:\:\alp...

0 votes

Prove the following : $ \cot^{-1} \bigg[ \large\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}} \bigg] = \frac{x}{2}, x \in \bigg( 0, \frac{\pi}{4} \bigg). $

answered Feb 28, 2013

Toolbox: Write $ sinx=2sin\large\frac{x}{2}\: cos\large\frac{x}{2}$ \( 1=sin^2\lar...

0 votes

Prove the following : $ \tan^{-1} \bigg(\large\frac{1}{2} \bigg) + \tan^{-1} \bigg( \large\frac{1}{5} \bigg) + \tan^{-1} \bigg( \frac{1}{8} \bigg)=\large\frac{\pi}{4} $

answered Feb 28, 2013

Toolbox: $ tan^{-1}x+tan^{-1}y=tan^{-1}\bigg( \large\frac{x+y}{1-xy} \bigg) , xy < 1$...

0 votes

What is the value of: \[tan^{-1} \bigg( \frac {x}{y} \bigg) -tan^{-1} \frac {x-y}{x+y} \]

answered Feb 28, 2013

Toolbox: \( tan^{-1}\alpha-tan^{-1}\beta=tan^{-1}\large\frac{\alpha-\beta}{1+\alpha\beta}\...

0 votes

Solve the following \[2tan^{-1} (cos x) = tan^{-1} (2\, cosec\, x) \]

answered Feb 28, 2013

Toolbox: $ 2tan^{-1}y=tan^{-1}\large\frac{2y}{1-y^2}\:\:|y|<1$ \(1-cos^2x=sin^2...

0 votes

Solve the following equation for x : $ \tan^{-1} \bigg( \large\frac{1-x}{1+x} \bigg) = \large\frac{1}{2}\tan^{-1}(x), x > 0 $

answered Feb 28, 2013

Toolbox:$tan(x-y)=\frac{tanx-tany}{1+tanx.tany}$$tan\frac{\pi}{4}=1$Let \(x=tan\theta,\:\Rightar...

0 votes

Prove the following : $\large\frac{9\pi}{8}-\frac{9}{4}sin^{-1}\bigg(\large\frac{1}{3} \bigg) = \frac{9}{4}sin^{-1}\bigg(\large \frac{2\sqrt 2}{3} \bigg) $

answered Feb 28, 2013

Toolbox: $ sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt{1-y^2}+y\sqrt{1-x^2} \bigg]$ \( sin^{-...

0 votes

Prove that $ \tan^{-1} \large\frac{63}{16}=\sin^{-1}\bigg( \large\frac{5}{13} \bigg) + \cos^{-1} \bigg( \large\frac{3}{5} \bigg). $

answered Feb 28, 2013

Toolbox:$ sin^{-1}x=tan^{-1}\large \frac{x}{\sqrt{1-x^2}}$\( cos^{-1}x=tan^{-1}\large \frac{\sqrt{...

0 votes

What is the principal value of $ \cos^{-1} \bigg( -\large\frac{\sqrt 3}{2} \bigg). $

answered Feb 28, 2013

Toolbox:$cos(\pi-\theta)=-cos\theta$We know that $cos\frac{\pi}{6}=\frac{\sqrt3}{2}$\(\Rightarro...

0 votes

Prove that $ 2sin^{-1}\frac{3}{5}-tan^{-1}\frac{17}{31}=\frac{\pi}{4} $

answered Feb 28, 2013

Toolbox:$ sin^{-1}x=tan^{-1}\large\frac{x}{\sqrt{1-x^2}}$ \( 2tan^{-1}x=tan^{-1}\large\frac{2x}...

0 votes

Find the values of x which satisfy the equation \[ sin^{-1}x+sin^{-1}(1-x)=cos^{-1}x.\]

answered Feb 28, 2013

Toolbox:$ sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt{1-y^2}+y\sqrt{1-x^2} \bigg]$\( cos^{-1}x=sin^...

0 votes

Solve the equation $ sin^{-1}6x+sin^{-1}6\sqrt 3 x = -\frac{\pi}{2}. $

answered Feb 28, 2013

Toolbox:$ sin^{-1}x+sin^{-1}y=sin^{-1}\: x\sqrt{1-y^2}+y\sqrt{1-x^2}$$sin(-\frac{\pi}{2})=-1$By ...

0 votes

Find principal value of the expression $ cos^{-1}(cos(-680^{\circ})).$

answered Feb 28, 2013

$680^{\circ}$ is not in the principal interval of cos which is $[0,180^{\circ}]$ Theref...

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 28, 2013

Toolbox: $\large\frac{1-tan^2\theta}{1+tan^2\theta}$$=cos2\theta$ Given $tan^{-1} \sqrt{x} ...

0 votes

Prove that : \[ 2tan^{-1} \bigg( \frac{1}{2} \bigg) + tan^{-1} \bigg( \frac{1}{7} \bigg) = tan^{-1} \bigg( \frac{31}{17} \bigg) \]

answered Feb 28, 2013

Toolbox:$ 2tan^{-1}x=tan^{-1}\frac{2x}{1-x^2}$\( tan^{-1}x+tan^{-1}y=tan^{-1}\frac{x+y}{1-xy}, xy ...

0 votes

Solve the following for x : \[ tan^{-1} \bigg( \frac{x-1}{x-2} \bigg) + tan^{-1} \bigg( \frac{x+1}{x+2} \bigg) = \frac{\pi}{4} \]

answered Feb 28, 2013

Toolbox: $tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}\: xy < 1$ tan\(\large\frac{\...

0 votes

Prove that $\bigg( tan \bigg( \frac{\pi}{4}+\frac{1}{2} cos^{-1}\frac{a}{b} \bigg)+tan \bigg(\frac{\pi}{4}-\frac{1}{2} cos^{-1}\frac{a}{b} \bigg) = \frac{2b}{a} \bigg)$

answered Feb 28, 2013

Toolbox:Take $ cos^{-1}\large\frac{a}{b}=\theta$$ \Rightarrow\:$ \( cos\theta=\large\frac{a}{b}\...

0 votes

Solve : \[ tan^{-1}(x+1)+tan^{-1}(x-1)=tan^{-1}\frac{8}{31}\]

answered Feb 28, 2013

Toolbox: $ tan^{-1}x+tan^{-1}y=tan^{-1}\bigg( \large\frac{x+y}{1-xy} \bigg) $By taking x+1 in t...

0 votes

Choose the correct answer. If $\sin^{-1}(1-x) -2 \sin^{-1} x=\large\frac {\pi}{2}, $ then $x$ is equal to

answered Feb 23, 2013

Toolbox: $ sin\bigg(\large \frac{\pi}{2}+y \bigg) = cosy$ \( cos2\theta = 1-2sin^2...

0 votes

Choose the correct answer $\sin (\tan^{-1}x), |x| <1 $ is equal to

answered Feb 23, 2013

Toolbox:If $\tan \theta = x$ then $\sin \theta = \large \frac {x}{\sqrt{(1+x^2)}}$Ans: (D) $ \frac{x...

0 votes

Solve the following $tan^{-1} \frac{1-x}{1+x} =\frac{1}{2} tan^{-1} x,(x>0)$

answered Feb 23, 2013

Toolbox: $ 2tan^{-1}y=tan^{-1}\large\frac{2y}{1-y^2}\:\:|y|<1$ Given $tan^{-1...

0 votes

Prove that \[\frac{9 \pi}{8} - \frac{9}{4} sin ^{-1}\frac{1}{3}= \frac{9}{4} sin^{-1}\frac{2\sqrt 2}{3}\]

answered Feb 23, 2013

Toolbox: $ sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt{1-y^2}+y\sqrt{1-x^2} \bigg]$ ...

0 votes

prove that \[tan^{-1} \left ( \frac {{\sqrt {1+x}}-{\sqrt {1-x}}}{{\sqrt {1+x}}+{\sqrt {1-x}}} \right ) = \frac {\pi}{4} -\frac{1}{2} cos^{-1}x,-\frac{1}{\sqrt2} \leq x \leq 1 \]

answered Feb 23, 2013

Toolbox: $ 1+cos\theta = 2cos^2\large\frac{\theta}{2} $ $ 1-cos\theta=2sin^2\large\f...

Ask Questions, Get Answers