Email
Chat with tutors
logo

Ask Questions, Get Answers

X
 

Answers posted by thanvigandhi_1

3013
answers
0
best answers
0 votes
0 votes
answered Feb 23, 2013
Toolbox: \(\large\frac{1-tan^2\theta}{1+tan^2\theta}\)\(=cos2\theta\) Given $tan^{...
0 votes
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
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 votes
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
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
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
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
answered Feb 23, 2013
Toolbox:Principal interval of tan is $(\large-\frac{\pi}{2},\large\frac{\pi}{2})$$\tan(\pi+x)=\tan x...
0 votes
answered Feb 23, 2013
Toolbox:Principal interval of cos is $[0,\pi]$$\cos(2\pi+x)=\cos x$Since $\large\frac{13\pi}{6}$ is ...
0 votes
answered Feb 23, 2013
Toolbox:$\tan^{-1}\sqrt{3}=\large\frac{\pi}{3}$$\cot^{-1}-\sqrt{3}=\pi-\large\frac{\pi}{6}$ Ans (B...
0 votes
answered Feb 23, 2013
Toolbox:$\sin^{-1}(-\large\frac{1}{2})=-\large\frac{\pi}{6}$$\sin\large\frac{\pi}{2}=1$Given $\sin \...
0 votes
answered Feb 23, 2013
Toolbox:$\cos(\pi+x)=-\cos x$$\cos(\pi+x)=cos(\pi-x)$Principal interval of cos is [0,\(\pi\)]Given $...
0 votes
answered Feb 23, 2013
Toolbox:\(tan(\pi-x)=-tanx\)\(tan^{-1}(-1)=-\frac{\pi}{4}\)Given $tan^{-1} \bigg( tan \frac{3\pi}{4}...
0 votes
answered Feb 23, 2013
Toolbox:$ \sin^{-1}x=\tan^{-1}\large \bigg(\frac{x}{\sqrt{1-x^2}}\bigg) $$\cot^{-1}x=\tan^{-1}\frac{...
0 votes
answered Feb 23, 2013
Toolbox:$sin(\pi - x) = sinx$The range of the principal value of $\sin^{-1}x$ is $\left [ -\frac{\pi...
0 votes
answered Feb 22, 2013
Toolbox:$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\large\frac{x+y}{1-xy}\: xy < 1$$\tan \large\frac{\pi}{4}...
0 votes
answered Feb 22, 2013
Toolbox: \( sin^{-1}x+cos^{-1}x=\large\frac{\pi}{2}\) \( sin\large\frac{\pi}{2}=1\: ...
0 votes
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
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
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
answered Feb 22, 2013
Toolbox:\( tan3\theta= \Large \frac{3tan\theta-tan^3\theta}{1-3tan^2\theta}\)$tan^{-1}tan\; \theta =...
0 votes
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
answered Feb 22, 2013
Toolbox:\( \large \frac{cosx-sinx}{cosx+sinx}\) \(=\large \frac{1-tanx}{1+tanx}\)\(tan(A-B)=\large \...
0 votes
answered Feb 22, 2013
Toolbox: \( 1-cos\theta = 2sin^2\frac{\theta}{2}\) \( 1+cos\theta = 2cos^2\frac{\the...
0 votes
answered Feb 22, 2013
Toolbox: \( sec^2\theta-1=tan^2\theta\) \( cot\theta = tan ( \large\frac{\pi}{2}-\th...
0 votes
answered Feb 22, 2013
Toolbox: \(cos3A=4cos^3A-3cosA\) \( cos^{-1}cosx=x \: \: \: if\: x \in [ 0\: \pi ] \...
0 votes
answered Feb 22, 2013
Toolbox: \( sin\: 3A=3\: sinA-4sin^3A\) \( sin^{-1}sinx=x\: \: \: \: if\: x \: \in\:...
0 votes
answered Feb 22, 2013
Ans : B$\frac{\pi}{3} - \bigg( \pi-\frac{\pi}{3} \bigg) = \frac{-\pi}{3} $
0 votes
answered Feb 22, 2013
Ans : \( \frac{\pi}{3}+2.\frac{\pi}{6}=\frac{2\pi}{3} \)
0 votes
answered Feb 22, 2013
Ans : \( \large\frac{\pi}{4}+\bigg( \pi-\large\frac{\pi}{3} \bigg)-\large\frac{\pi}{6} = \large...
0 votes
answered Feb 22, 2013
Ans : $ cosec^{-1}cosec \bigg( \large\frac{-\pi}{4} \bigg) = \large\frac{-\pi}{4} $
0 votes
answered Feb 22, 2013
Ans : $ cos^{-1} \bigg( cos \bigg( \pi-\frac{\pi}{4} \bigg) \bigg) = \frac{3\pi}{4} $
0 votes
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
answered Feb 22, 2013
Ans : $ sec^{-1}sec\frac{\pi}{6}=\frac{\pi}{6} $
0 votes
answered Feb 22, 2013
Ans : $( tan^{-1}tan \frac{-\pi}{4}=\frac{-\pi}{4} )$
0 votes
answered Feb 22, 2013
Ans : $( cos^{-1} \bigg( cos \bigg( \pi - \large\frac{\pi}{3} \bigg) \bigg)=\large\frac{2 \pi}{3})$
0 votes
answered Feb 22, 2013
Ans : $tan^{-1}tan \large\frac{-\pi}{3}=\large\frac{-\pi}{3} $
0 votes
answered Feb 22, 2013
Ans : $ cos^{-1} \bigg( cos\large\frac{\pi}{6} \bigg) = \large\frac{\pi}{6} $
0 votes
answered Feb 22, 2013
Toolbox:Principal interval of $( \sin x) is \bigg[ \large\frac{\pi}{2}, \large\frac{\pi}{2} \bigg] $...
0 votes
answered Feb 20, 2013
Toolbox:$\cot^{-1}x=\tan^{-1}\large\frac{1}{x} $$ 2\tan^{-1}x=\tan^{-1}\large\frac{2x}{1-x^2} $$\tan...
0 votes
answered Feb 20, 2013
Toolbox:$(cos(\pi-\theta)=-cos\theta)$Principal interval of cos is $( [0,\pi])$$(cos\large\frac{\pi}...
0 votes
answered Feb 20, 2013
Toolbox: \( tan^{-1}1=\large\frac{\pi}{4} \) \( tan^{-1}x+tan^{-1}y=tan^{-1}\large\f...
0 votes
answered Feb 20, 2013
Toolbox: \( sin^{-1}x+cos^{-1}x=\large\frac{\pi}{2}\) \( sin\large\frac{\pi}{2}=1\: ...
0 votes
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
answered Feb 20, 2013
Toolbox: \( tan^{-1}x+tan^{-1}y=tan^{-1}\large\frac{x+y}{1-xy}\:\:xy<1 \) \(tan^{...
0 votes
answered Feb 20, 2013
Toolbox: \(sin^{-1}x+cos^{-1}x=\large\frac{\pi}{2}\) for all x in the domain  ...
0 votes
answered Feb 20, 2013
Toolbox: \( cos^{-1}x = tan^{-1} \large\frac{\sqrt{1-x^2}}{x} \) \( sincos^{-1}x=\sq...
0 votes
answered Feb 20, 2013
Toolbox: \( sin^{-1}x=cos^{-1}\sqrt{1-x^2} \) \(cos\theta=sin(\large\frac{\pi}{2}-\t...
0 votes
answered Feb 20, 2013
Toolbox:Principal interval of $\sin^{-1}x \;is\;[-\large\frac{\pi}{2},\frac{\pi}{2}]$$\sin\large\fra...
Home Ask Tuition Questions
Your payment for is successful.
Continue
...