જો theta = 2 tan^{-1} frac{1}{8} + 2 tan^{-1} | vedclass

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(B) આપેલ છે,$	heta = 2 	an^{-1} \frac{1}{8} + 2 	an^{-1} \frac{1}{5} + 	an^{-1} \frac{1}{7}$.સૂત્ર $2 	an^{-1} x = 	an^{-1} \frac{2x}{1-x^2}$ નો

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(B) આપેલ છે,$	heta = 2 	an^{-1} \frac{1}{8} + 2 	an^{-1} \frac{1}{5} + 	an^{-1} \frac{1}{7}$.સૂત્ર $2 	an^{-1} x = 	an^{-1} \frac{2x}{1-x^2}$ નો ઉપયોગ કરતા:$2 	an^{-1} \frac{1}{8} = 	an^{-1} \frac{2/8}{1-1/64} = 	an^{-1} \frac{16}{63}$.$2 	an^{-1} \frac{1}{5} = 	an^{-1} \frac{2/5}{1-1/25} = 	an^{-1} \frac{5}{12}$.હવે,$	heta = 	an^{-1} \frac{16}{63} + 	an^{-1} \frac{5}{12} + 	an^{-1} \frac{1}{7}$.$	an^{-1} x + 	an^{-1} y = 	an^{-1} \frac{x+y}{1-xy}$ નો ઉપયોગ કરતા:$	an^{-1} \frac{16}{63} + 	an^{-1} \frac{5}{12} = 	an^{-1} \frac{3}{4}$.તેથી,$	heta = 	an^{-1} \frac{3}{4} + 	an^{-1} \frac{1}{7} = 	an^{-1} (1) = \frac{\pi}{4}$.આમ,$	an \frac{	heta}{2} = 	an \frac{\pi}{8} = \sqrt{2} - 1$.$\sqrt{m} + \sqrt{n} = -1 + \sqrt{2}$ અને $m તેથી $(m^n + n^m)^{m+n} = (1^2 + 2^1)^{1+2} = 3^3 = 27$.

યાદી-$I$	યાદી-$II$
$A$. $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\sin ^{-1} \frac{1}{3}$	$I$. $k \pi \pm(-1)^k \frac{\pi}{6}, k \in Z$
$B$. $\sin ^{-1}\left(\frac{(-1)^n}{2}\right), n \in Z$	$II$. $k \pi \pm 1, k \in Z$
$C$. $\tan ^{-1}\left(\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right)$	$III$. $\frac{3}{2}$
$D$. $\sin ^{-1}\|\sin x\|=\sqrt{\sin ^{-1}\|\sin x\|} \Rightarrow x \in$	$IV$. $\frac{3 \pi}{8}$
	$V$. $\frac{\pi}{2}$

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