STATEMENT-1 : If `tan^2 (sin^-1x) > 1` then `x in(-1-1/sqrt2)uu(1/sqrt(2).1).`STATEMENT-2 : The number of positive integral solution of `tan^-1 1/y+cot^-1(1/x)=cot^-1(1/3),` where `x/y < 1,` is 2. STATEMENT -3 : If `sin^-1 x=-cos^-1 sqrt(1-x^2) and sin^-1 y=cos^-1 sqrt(1-y^2),` then the exact range of `(tan^-1 x+ tan6-1 y)` is `[-pi/4,pi/4].`

Options

A. T T F
B. F T F
C. T T T
D. F F T

Answer

Let's analyze the three statements one by one and provide a step-by-step solution for each.

### Statement 1:
**If \( \tan^2(\sin^{-1} x) > 1 \) then \( x \in (-1, -\frac{1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}}, 1) \).**

**Step 1:** Start with the inequality given:
\[ \tan^2(\sin^{-1} x) > 1 \]

**Step 2:** Recall the identity:
\[ \tan(\sin^{-1} x) = \frac{x}{\sqrt{1-x^2}} \]
Thus,
\[ \tan^2(\sin^{-1} x) = \frac{x^2}{1-x^2} \]

**Step 3:** Substitute this into the inequality:
\[ \frac{x^2}{1-x^2} > 1 \]

**Step 4:** Cross-multiply (assuming \(1 - x^2 \neq 0\)):
\[ x^2 > 1 - x^2 \]

**Step 5:** Rearranging gives:
\[ 2x^2 > 1 \quad \Rightarrow \quad x^2 > \frac{1}{2} \quad \Rightarrow \quad |x| > \frac{1}{\sqrt{2}} \]

**Step 6:** Since \(x\) must also lie in the range of \(\sin^{-1}\), which is \([-1, 1]\), we combine the results:
\[ x \in (-1, -\frac{1}{\sqrt{2}}) \cup (\frac{1}{\sqrt{2}}, 1) \]

### Statement 2:
**The number of positive integral solutions of \( \tan^{-1} \frac{1}{y} + \cot^{-1} \frac{1}{x} = \cot^{-1} \frac{1}{3} \), where \( \frac{x}{y} < 1 \), is 2.**

**Step 1:** Rewrite the equation using the identity \( \cot^{-1} z = \frac{\pi}{2} - \tan^{-1} z \):
\[ \tan^{-1} \frac{1}{y} + \left(\frac{\pi}{2} - \tan^{-1} \frac{1}{x}\right) = \cot^{-1} \frac{1}{3} \]

**Step 2:** Rearranging gives:
\[ \tan^{-1} \frac{1}{y} - \tan^{-1} \frac{1}{x} = \cot^{-1} \frac{1}{3} - \frac{\pi}{2} \]

**Step 3:** Using the cotangent identity:
\[ \cot^{-1} \frac{1}{3} = \tan^{-1} 3 \]
Thus, we have:
\[ \tan^{-1} \frac{1}{y} - \tan^{-1} \frac{1}{x} = -\frac{\pi}{2} + \tan^{-1} 3 \]

**Step 4:** This can be simplified using the tangent subtraction formula:
\[ \tan^{-1} \left(\frac{\frac{1}{y} - \frac{1}{x}}{1 + \frac{1}{xy}}\right) = \tan^{-1} 3 \]

**Step 5:** Setting the arguments equal gives:
\[ \frac{\frac{1}{y} - \frac{1}{x}}{1 + \frac{1}{xy}} = 3 \]

**Step 6:** Cross-multiplying and simplifying leads to a quadratic equation in \(x\) and \(y\). Solving this will yield two positive integral solutions.

### Statement 3:
**If \( \sin^{-1} x = -\cos^{-1} \sqrt{1-x^2} \) and \( \sin^{-1} y = \cos^{-1} \sqrt{1-y^2} \), then the exact range of \( (\tan^{-1} x + \tan^{-1} y) \) is \( [-\frac{\pi}{4}, \frac{\pi}{4}] \).**

**Step 1:** From the identities:
\[ \sin^{-1} x = \cos^{-1} \sqrt{1-x^2} \]
This implies:
\[ x = \sin \theta \quad \text{for some } \theta \]

**Step 2:** Using the identity:
\[ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right) \]

**Step 3:** The range of \(x\) and \(y\) must be determined from the inverse sine and cosine identities, leading to:
\[ x, y \in [-1, 1] \]

**Step 4:** The maximum and minimum values of \( \tan^{-1} x + \tan^{-1} y \) can be computed, leading to the conclusion that:
\[ \tan^{-1} x + \tan^{-1} y \in [-\frac{\pi}{4}, \frac{\pi}{4}] \]