If a, b, c are positive real numbers and the | vedclass

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

Question

(A) Let $s^2 = \frac{a+b+c}{abc}$. Then the expression becomes $	heta = 	an^{-1}\sqrt{a^2 s^2} + 	an^{-1}\sqrt{b^2 s^2} + 	an^{-1}\sqrt{c^2 s^2}$.

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Let $s^2 = \frac{a+b+c}{abc}$. Then the expression becomes $	heta = 	an^{-1}\sqrt{a^2 s^2} + 	an^{-1}\sqrt{b^2 s^2} + 	an^{-1}\sqrt{c^2 s^2}$.Since $a, b, c$ are positive,$	heta = 	an^{-1}(as) + 	an^{-1}(bs) + 	an^{-1}(cs)$.Using the formula $	an^{-1} x + 	an^{-1} y + 	an^{-1} z = 	an^{-1}\left(\frac{x+y+z-xyz}{1-(xy+yz+zx)}ight)$:$	an 	heta = \frac{as+bs+cs - (as)(bs)(cs)}{1 - (asbs + bscs + csas)} = \frac{s(a+b+c) - s^3(abc)}{1 - s^2(ab+bc+ca)}$.Substitute $s^2 = \frac{a+b+c}{abc}$,which implies $s^2(abc) = a+b+c$:$	an 	heta = \frac{s(a+b+c) - s(s^2 abc)}{1 - s^2(ab+bc+ca)} = \frac{s(a+b+c) - s(a+b+c)}{1 - s^2(ab+bc+ca)} = \frac{0}{1 - s^2(ab+bc+ca)} = 0$.Alternatively,let $a=b=c=1$. Then $	heta = 	an^{-1}\sqrt{3} + 	an^{-1}\sqrt{3} + 	an^{-1}\sqrt{3} = 3 	imes 60^\circ = 180^\circ = \pi$. Thus,$	an 	heta = 	an \pi = 0$.

If $a, b, c$ are positive real numbers and the value of $\theta = \tan^{-1}\sqrt{\frac{a(a+b+c)}{bc}} + \tan^{-1}\sqrt{\frac{b(a+b+c)}{ca}} + \tan^{-1}\sqrt{\frac{c(a+b+c)}{ab}}$,then $\tan \theta$ is equal to

Similar Questions

If the value of $x$ satisfying the equation $\sin^{-1} \sqrt{1-x^2} = \tan^{-1} \sqrt{\frac{2}{x}-1}$ is $\frac{a}{b}$ (where $a$ and $b$ are coprime),then the value of $a^2 + b^2$ is

If $\tan^{-1}\left(\frac{x-1}{x+1}\right) + \tan^{-1}\left(\frac{2x-1}{2x+1}\right) = \tan^{-1}\left(\frac{23}{36}\right)$,then $x =$

If $\sin^{-1}(\tan \frac{\pi}{4}) - \sin^{-1}(\sqrt{\frac{3}{x}}) = \frac{\pi}{6}$, then $x$ is a root of the equation:

The number of solutions of the equation $2\tan^{-1}(\cos^2 x) = \tan^{-1}(2\csc^2 x)$ in the interval $[0, 5\pi]$ is $m$. Then which of the following is true?

If $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$, then $x \in$

Vedclass Test Series

Exam Paper Generator

Online Exam Module