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The number of real roots of the equation \tan ^{-1} \sqrt{\mathrm{x}(\mathrm{x}+1)}+\sin ^{-1} \sqrt{\mathrm{x}^{2}+\mathrm{x}+1}=\frac{\pi}{4} is :
 
Option: 1 1
Option: 2 2
Option: 3 4  
Option: 4 0

\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{4} .

Domain

\begin{aligned} &1, \quad x(x+1) \geqslant 0 \\ &\Rightarrow \quad x \in(-\infty,-1] \cup[0, \infty) \\ &2 \cdot \quad x^{2}+x+1 \geqslant 0 \Rightarrow x \in R \end{aligned}

\begin{aligned} \\&\text { 3. } \sqrt{x^{2}+x+1} \leq 1 \\ \Rightarrow & x^{2}+x+1 \leq 1 \\ \Rightarrow & x^{2}+x \leq 0 \\ \Rightarrow & x(x+1) \leq 0 \\ \Rightarrow & x \in[-1,0] \end{aligned}

Overall domain is intersection of these 3

\Rightarrow \text { domain }=\{-1,0\} \text {. }

 

For x= -1

Equation is \tan ^{-1}(0)+\sin ^{-1}(1)=\frac{\pi}{4} \Rightarrow \frac{\pi}{2} = \frac{\pi}{4}

So x= -1 do not satisfy it

 

For x= 0

Equation is \tan ^{-1}(0)+\sin ^{-1}(1)=\frac{\pi}{4} \Rightarrow \frac{\pi}{2} = \frac{\pi}{4}

x= 0 also does not satisfy it

So, no solution

Hence, the correct answer is option (4)

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Posted by

Kuldeep Maurya

Solve the equation x:\sin ^{-1}\left ( \frac{5}{x} \right )+\sin ^{-1}\left ( \frac{12}{x} \right )=\frac{\pi}{2}(x\neq 0)

 

 
 
 
 
 

\\\sin ^{-1}\left ( \frac{5}{x} \right )+\sin ^{-1}\left ( \frac{12}{x} \right )=\frac{\pi}{2}

\\ \sin ^{-1}\left(\frac{5}{x}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{12}{x}\right) \\ \sin ^{-1}\left(\frac{5}{x}\right)=\cos ^{-1}\left(\frac{12}{x}\right) \\ \cos ^{-1}\left(\frac{\sqrt{x^{2}-25}}{x}\right)=\cos ^{-1}\left(\frac{12}{x}\right) \\ \left(\frac{\sqrt{x^{2}-25}}{x}\right)=\left(\frac{12}{x}\right) \\ \sqrt{x^{2}-25}=12 \\ x^{2}-25=144 \\ x^{2}=144+25=169 \\ x=13

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Posted by

Safeer PP

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Prove that \tan \left [ 2\tan ^{-1}\left ( \frac{1}{2} \right ) - \cot ^{-1}3\right ]=\frac{9}{13}

 

 
 
 
 
 

To prove:

\tan \left [ 2\tan ^{-1}\left ( \frac{1}{2} \right ) - \cot ^{-1}3\right ]=\frac{9}{13}

\\LHS = \tan \left[2 \tan ^{-1} \frac{1}{2}-\cot ^{-1} 3\right] \\ =\tan \left[\tan ^{-1}\left(\frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}\right)-\cot ^{-1} 3\right] \\ =\tan \left[\tan ^{-1}\left\{\frac{1}{1-\frac{1}{4}}\right\}-\tan ^{-1} \frac{1}{3}\right] \\ =\tan \left[\tan ^{-1} \frac{4}{3}-\tan ^{-1} \frac{1}{3}\right]

\\ =\tan \left[\tan ^{-1}\left(\frac{\frac{4}{3}-\frac{1}{3}}{1+\frac{4}{9}}\right)\right] \\ =\tan \left[\tan ^{-1}\left(\frac{\frac{4-1}{3}}{\frac{9+4}{9}}\right)\right] \\ =\tan \left[\tan ^{-1}\left(\frac{1}{\frac{13}{9}}\right)\right] \\ =\tan \left[\tan ^{-1}\left(\frac{9}{13}\right)\right] \\ =\frac{9}{13} = RHS

Hence proved 

 

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Posted by

Safeer PP

Using differential, find the approximate value of \sqrt{36.6} upto 2 decimal places.

 

f(x) = \sqrt{x}

f(36.6) = \sqrt{36.6}

f(36.6) = f(36) + (0.6) f'(x)

f(36.6) = 6 + (0.6) \frac{1}{2\sqrt{36}}

f(36.6) = 6 + 0.05 = 6.05

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Posted by

Safeer PP

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Evaluate : \sin \left [ \frac{\pi}{3}-\sin ^{-1}\left ( -\frac{1}{2} \right ) \right ]

 

 

\\\sin \left [ \frac{\pi}{3}-\sin ^{-1}\left ( -\frac{1}{2} \right ) \right ] \\\\ = \sin \left [ \frac{\pi}{3}-(-\frac{\pi}{6})\right ] \\\\ = \sin \left [ \frac{\pi}{3} +\frac{\pi}{6})\right ] \\\\ = \sin \left [ \frac{\pi}{2}\right ] \\\\ = 1

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Safeer PP

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

 

\\ \sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2} \\ \\ \sin^{-1} (1-x)=\frac{\pi}{2}+2 \sin ^{-1} x \\\\ (1-x)=\sin \left(\frac{\pi}{2}+2 \sin ^{-1} x\right) \\\\ $ $(1-x)=\cos \left(2 \sin ^{-1} x\right) \\\\ 1-x=1-2 x^{2} \\ $ $ 2 x^{2}-x=0 \\ x=0, x=\frac{1}{2} \\ x=\frac{1}{2}$ does not satisty the given equation \\ x=0 is the only solution.

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Posted by

Safeer PP

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Find the value of \sin ^{-1}\left (-\frac{17 \pi}{8} \right )

 

\\ \sin ^{-1}\left[\sin \left(-\frac{17 \pi}{8}\right)\right] \\\\\ =-\sin ^{-1}\left[\sin \left(2 \pi+\frac{\pi}{8}\right)\right] \\ =-\frac{\pi}{8}

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Posted by

Safeer PP

Prove that :

\frac{9 \pi}{8}-\frac{9}{4}\sin^{-1}\left ( \frac{1}{3} \right )=\frac{9}{4}\sin ^{-1}\left ( \frac{2\sqrt{2}}{3} \right )

 

 

 
 
 
 
 

To Prove:

\frac{9 \pi}{8}-\frac{9}{4}\sin^{-1}\left ( \frac{1}{3} \right )=\frac{9}{4}\sin ^{-1}\left ( \frac{2\sqrt{2}}{3} \right )

\frac{9 \pi}{8}=\frac{9}{4}\sin^{-1}\left ( \frac{1}{3} \right )+\frac{9}{4}\sin ^{-1}\left ( \frac{2\sqrt{2}}{3} \right )

\sin^{-1}\left ( \frac{1}{3} \right )+\sin ^{-1}\left ( \frac{2\sqrt{2}}{3} \right ) = \frac{ \pi}{2}

\sin ^{-1}\left ( \frac{2\sqrt{2}}{3} \right ) = \frac{ \pi}{2} - \sin^{-1}\left ( \frac{1}{3} \right )

\sin ^{-1}\left ( \frac{2\sqrt{2}}{3} \right ) = \cos^{-1}\left ( \frac{1}{3} \right )

\sin ^{-1}\left ( \frac{2\sqrt{2}}{3} \right ) = \sin ^{-1}\left ( \sqrt{1-(\frac{1}{3})^2} \right )

\sin ^{-1}\left ( \frac{2\sqrt{2}}{3} \right ) = \sin ^{-1}\left (\frac{2\sqrt{2}}{3} \right )

Hence proved.

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Posted by

Safeer PP

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Express\tan ^{-1}\left [ \frac{\cos x}{1-\sin x} \right ],-\frac{3\pi }{2}<x<\frac{\pi}{2} in the simplest form.

 

 

Given:

\tan ^{-1}\left [ \frac{\cos x}{1-\sin x} \right ],-\frac{3\pi }{2}<x<\frac{\pi}{2}

\\ \tan ^{-1}\left(\frac{\cos x}{1-\sin x}\right) \\\\=\tan ^{-1}\left(\frac{\sin \left(\frac{\pi}{2}-x\right)}{1-\cos \left(\frac{\pi}{2}-x\right)}\right) \\\\=\tan ^{-1}\left(\frac{2\sin \left(\frac{\pi}{4}-\frac{x}{2}\right)\cos \left(\frac{\pi}{4}-\frac{x}{2}\right)}{2\sin^2 \left(\frac{\pi}{4}-\frac{x}{2}\right)}\right)

\\ =\tan ^{-1}\left[\cot \left(\frac{\pi}{4}-\frac{x}{2}\right)\right] \\\\ =\tan ^{-1}\left[\tan \left(\frac{\pi}{2}-\frac{\pi}{4}+\frac{x}{2}\right)\right]\\\\=\frac{\pi}{4}+\frac{x}{2}

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Posted by

Safeer PP

Solve for x :

sin^{-1} 4x + sin^{-1} 3x = - \frac{\pi}{2}

 

 

 

 
 
 
 
 

Given:

\sin^{-1} 4x + \sin^{-1} 3x = - \frac{\pi}{2}

\sin ^{-1}(4 x)+\sin ^{-1}(3 x)=-\frac{\pi}{2}

\sin ^{-1}(3 x)=-\frac{\pi}{2}-\sin ^{-1}(4 x)

3 x=\sin \left(-\frac{\pi}{2}-\sin ^{-1} 4 x\right)

3 x=-\sin \left(\frac{\pi}{2}+\sin ^{-1} 4 x\right)

3x=-\cos \left(\sin ^{-1} 4 x\right)

-3 x=\sqrt{1-4 x^{2}}

\quad 9 x^{2}=1-16 x^{2}

25 x^{2}=1 \\

x^{2}=\frac{1}{25}

x=\pm \frac{1}{5}

$At $ x =\frac{1}{5}

\sin ^{-1} 4 x+\sin ^{-1} 3 x>0 \Rightarrow x \neq \frac{1}{5}

x=-\frac{1}{5} \text { is the only solution }

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Posted by

Safeer PP

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