If sin ^{-1}left(x-frac{x^2}{2}+frac{x^3}{4}- | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given equation is $\sin ^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\ldots \infty\right) + \cos ^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\ldots

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The given equation is $\sin ^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\ldots \inftyight) + \cos ^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\ldots \inftyight)=\frac{\pi}{2}$.Both series are infinite geometric progressions with common ratios $-x/2$ and $-x^2/2$ respectively.The sum of an infinite $G$.$P$. is $S = \frac{a}{1-r}$.For the first series,$a=x$ and $r=-x/2$,so the sum is $\frac{x}{1-(-x/2)} = \frac{x}{1+x/2} = \frac{2x}{2+x}$.Wait,the series given is $x - x^2/2 + x^3/4 - \dots$,which is a $G$.$P$. with $a=x$ and $r=-x/2$. The sum is $\frac{x}{1-(-x/2)} = \frac{2x}{2+x}$.Similarly,for the second series $x^2 - x^4/2 + x^6/4 - \dots$,$a=x^2$ and $r=-x^2/2$. The sum is $\frac{x^2}{1-(-x^2/2)} = \frac{2x^2}{2+x^2}$.Using the property $\sin ^{-1}(u) + \cos ^{-1}(v) = \pi/2$,we must have $u=v$.Thus,$\frac{2x}{2+x} = \frac{2x^2}{2+x^2}$.Since $x 
eq 0$,we divide by $2x$: $\frac{1}{2+x} = \frac{x}{2+x^2}$.$2+x^2 = x(2+x) \implies 2+x^2 = 2x+x^2$.$2x = 2 \implies x=1$.

If $\sin ^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\ldots \infty\right) + \cos ^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\ldots \infty\right)=\frac{\pi}{2}$ and $0 < x < \sqrt{2}$,then $x$ is equal to

Similar Questions

If $\tan ^{-1}\left(\frac{1}{3}\right) + \tan ^{-1}\left(\frac{1}{7}\right) + \tan ^{-1}\left(\frac{1}{13}\right) + \tan ^{-1}\left(\frac{1}{21}\right) + \tan ^{-1}\left(\frac{1}{31}\right) = \tan ^{-1}\left(\frac{p}{q}\right)$,where $p$ and $q$ are relatively prime numbers,then $p + q$ is equal to-

Evaluate: ${\tan ^{ - 1}}\left[ {\frac{{\cos x}}{{1 + \sin x}}} \right]$

If $3{\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} - 4{\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}} + 2{\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} = \frac{\pi }{3}$,then $x$ =

$\cot ^{-1}\left(2 \cdot 1^2\right)+\cot ^{-1}\left(2 \cdot 2^2\right)+\cot ^{-1}\left(2 \cdot 3^2\right)+\ldots \ldots \ldots \infty =$

$\cot ^{-1}\left(\frac{1}{\sqrt{x^2-1}}\right)=$ . . . . . . where,$x>1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module