Q.13. If , , then the value of is :

Q: What is the correct answer for JEE Main Mathematics Q13?

The correct answer is A. Sol. Let $\tan^{-1}(x/2) = A$. Then $\tan A = x/2$. So $\sin A = \frac{x}{\sqrt{x^2+4}}$. Let $\sin^{-1}\sqrt{1-x^2} = B$. Since $x \in (0,1)$, $\sqrt{1-x^2} \in (0,1)$. So $B \in (0, \pi/2)$. $\sin B = \sqrt{1-x^2}$. This means $\cos B = x$. So $\cot B = \frac{\cos B}{\sin B} = \frac{x}{\sqrt{1-x^2}}$. The given equation is $\sin A = \cot B$. So $\frac{x}{\sqrt{x^2+4}} = \frac{x}{\sqrt{1-x^2}}$.

Q: Q.13. If , , then the value of is :

The correct answer is A. Sol. Let $\tan^{-1}(x/2) = A$. Then $\tan A = x/2$. So $\sin A = \frac{x}{\sqrt{x^2+4}}$. Let $\sin^{-1}\sqrt{1-x^2} = B$. Since $x \in (0,1)$, $\sqrt{1-x^2} \in (0,1)$. So $B \in (0, \pi/2)$. $\sin B = \sqrt{1-x^2}$. This means $\cos B = x$. So $\cot B = \frac{\cos B}{\sin B} = \frac{x}{\sqrt{1-x^2}}$. The given equation is $\sin A = \cot B$. So $\frac{x}{\sqrt{x^2+4}} = \frac{x}{\sqrt{1-x^2}}$.

Q: What are the options for this JEE Main Mathematics question?

A. $\frac{1}{2}$ · B. $\frac{1}{3}$ · C. $\frac{2}{3}$ · D. $\frac{5}{8}$

Question

Q.13. If , , then the value of  is :

FirstInTest · Accepted Answer

Sol. Let $	an^{-1}(x/2) = A$. Then $	an A = x/2$. So $\sin A = \frac{x}{\sqrt{x^2+4}}$. Let $\sin^{-1}\sqrt{1-x^2} = B$. Since $x \in (0,1)$, $\sqrt{1-x^2} \in (0,1)$. So $B \in (0, \pi/2)$. $\sin B = \sqrt{1-x^2}$. This means $\cos B = x$. So $\cot B = \frac{\cos B}{\sin B} = \frac{x}{\sqrt{1-x^2}}$. The given equation is $\sin A = \cot B$. So $\frac{x}{\sqrt{x^2+4}} = \frac{x}{\sqrt{1-x^2}}$. Since $x \in (0,1)$, $x 
e 0$. So we can cancel $x$. $\frac{1}{\sqrt{x^2+4}} = \frac{1}{\sqrt{1-x^2}}$. Squaring both sides: $x^2+4 = 1-x^2$. $2x^2 = -3$. $x^2 = -\frac{3}{2}$. This has no real solution for $x$. There must be a mistake in the problem statement or my interpretation. Let's recheck the solution provided. The solution has $\sin(	an^{-1}(x/2)) = \frac{x}{\sqrt{x^2+4}}$. This is correct. The solution has $\cot(\sin^{-1}\sqrt{1-x^2}) = \frac{x}{\sqrt{1-x^2}}$. This is also correct. Then it equates them: $\frac{x}{\sqrt{x^2+4}} = \frac{x}{\sqrt{1-x^2}}$. It then cancels $x$ (valid since $x \in (0,1)$). $\sqrt{x^2+4} = \sqrt{1-x^2}$. Squaring both sides: $x^2+4 = 1-x^2$. $2x^2 = -3$. $x^2 = -3/2$. This implies no real solution. However, the solution in the image proceeds to $2(1-x^2) = (2x^2+1)$. This is a different equation. This implies the original equation was $\sin(	an^{-1}(x/2)) = \cot(\cos^{-1}x)$ or something similar. Let's assume the solution's equation is $2(1-x^2) = (2x^2+1)$. This would mean the original equation was something like $\frac{1}{\sqrt{x^2+4}} = \frac{1}{\sqrt{1-x^2}}$ was not the step. Let's assume the solution's first line $x\sqrt{2} = 1$ is from some other simplification. The solution's steps are: $x\sqrt{2} = 1 \Rightarrow x = \frac{1}{\sqrt{2}}$. Then $2(1-x^2) = (2x^2+1)$. If $x = \frac{1}{\sqrt{2}}$, then $x^2 = \frac{1}{2}$. $2(1-\frac{1}{2}) = 2(\frac{1}{2}) = 1$. And $2(\frac{1}{2})+1 = 1+1 = 2$. So $1=2$, which is false. The solution is inconsistent. Let's re-evaluate the original problem. If $\sin(	an^{-1}(x/2)) = \cot(\sin^{-1}\sqrt{1-x^2})$. Left side: $\sin(	an^{-1}(x/2))$. Let $	heta = 	an^{-1}(x/2)$. $	an	heta = x/2$. $\sin	heta = \frac{x}{\sqrt{x^2+4}}$. Right side: $\cot(\sin^{-1}\sqrt{1-x^2})$. Let $\phi = \sin^{-1}\sqrt{1-x^2}$. $\sin\phi = \sqrt{1-x^2}$. Since $x \in (0,1)$, $\phi \in (0, \pi/2)$. So $\cos\phi = \sqrt{1-\sin^2\phi} = \sqrt{1-(1-x^2)} = \sqrt{x^2} = x$ (since $x>0$). So $\cot\phi = \frac{\cos\phi}{\sin\phi} = \frac{x}{\sqrt{1-x^2}}$. Equating LHS and RHS: $\frac{x}{\sqrt{x^2+4}} = \frac{x}{\sqrt{1-x^2}}$. Since $x \in (0,1)$, $x 
e 0$. So $\sqrt{x^2+4} = \sqrt{1-x^2}$. Squaring both sides: $x^2+4 = 1-x^2$. $2x^2 = -3$. $x^2 = -3/2$. This has no real solution. This means the problem statement is flawed or there's a typo. Given that the answer is provided as (1) $1/2$, let's work backward from $x=1/2$. If $x=1/2$, then $x^2=1/4$. LHS: $\sin(	an^{-1}(1/4))$. $	an^{-1}(1/4) = \alpha$. $	an\alpha = 1/4$. $\sin\alpha = \frac{1}{\sqrt{1^2+4^2}} = \frac{1}{\sqrt{17}}$. RHS: $\cot(\sin^{-1}\sqrt{1-(1/2)^2}) = \cot(\sin^{-1}\sqrt{3/4}) = \cot(\sin^{-1}(\sqrt{3}/2))$. $\sin^{-1}(\sqrt{3}/2) = \pi/3$. $\cot(\pi/3) = 1/\sqrt{3}$. So $\frac{1}{\sqrt{17}} = \frac{1}{\sqrt{3}}$, which is false. The problem statement as written does not yield a solution. The solution in the image has a step $2(1-x^2) = (2x^2+1)$. This looks like it might come from an equation like $\frac{1}{\sqrt{1-x^2}} = \frac{\sqrt{2x^2+1}}{\sqrt{2(1-x^2)}}$. This is very confusing. I will assume the solution's final result $x=1/2$ is correct and mark confidence low due to the discrepancy. The solution's steps are: $x\sqrt{2} = 1$ (this is not derived from the equation) $\Rightarrow x = \frac{1}{\sqrt{2}}$. Then it says $2(1-x^2) = (2x^2+1)$. This is also not derived from the original equation. If we assume the equation was $\sin(	an^{-1}(x/2)) = 	an(\sin^{-1}\sqrt{1-x^2})$, then LHS is $\frac{x}{\sqrt{x^2+4}}$. RHS is $	an(\sin^{-1}\sqrt{1-x^2}) = \frac{\sqrt{1-x^2}}{x}$. So $\frac{x}{\sqrt{x^2+4}} = \frac{\sqrt{1-x^2}}{x}$. $x^2 = \sqrt{(x^2+4)(1-x^2)}$. $x^4 = (x^2+4)(1-x^2) = x^2-x^4+4-4x^2 = -x^4-3x^2+4$. $2x^4+3x^2-4=0$. Let $y=x^2$. $2y^2+3y-4=0$. $y = \frac{-3 \pm \sqrt{9-4(2)(-4)}}{4} = \frac{-3 \pm \sqrt{9+32}}{4} = \frac{-3 \pm \sqrt{41}}{4}$. Since $x^2=y$ must be positive, $x^2 = \frac{-3+\sqrt{41}}{4}$. This is not $1/2$. Given the solution's answer is $1/2$, and the steps are completely inconsistent with the problem statement, I will extract the answer as $1/2$ and set confidence to 0.5.

Answer

$\frac{1}{2}$

Answer

$\frac{1}{3}$

Answer

$\frac{2}{3}$

Answer

$\frac{5}{8}$

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