Statement:1 `sinx =1/p rArr tanx =1/sqrt(p^(2)-1), p ne 1`.
Statement-2: `cosx = x^(3)+ 3/x, x gt 0` has no solution in R.
Statement-3: `tan^(2)x - sin^(2)x = tan^(2)x. sin^(2)x, x in R`

Options

A. FFF
B. FTF
C. TTT
D. TTF

Answer

To solve the given statements step by step, let's analyze each statement one by one.

### Statement 1:
**Given:** \( \sin x = \frac{1}{p} \) implies \( \tan x = \frac{1}{\sqrt{p^2 - 1}} \), where \( p \neq 1 \).

**Step 1:** Start with the given equation \( \sin x = \frac{1}{p} \).

**Step 2:** Recall the definition of sine in a right triangle:
\[ \sin x = \frac{\text{Perpendicular}}{\text{Hypotenuse}}. \]
Here, we can identify the perpendicular as 1 and the hypotenuse as \( p \).

**Step 3:** Use the Pythagorean theorem to find the base:
\[ \text{Base} = \sqrt{\text{Hypotenuse}^2 - \text{Perpendicular}^2} = \sqrt{p^2 - 1}. \]

**Step 4:** Now, calculate \( \tan x \):
\[ \tan x = \frac{\text{Perpendicular}}{\text{Base}} = \frac{1}{\sqrt{p^2 - 1}}. \]

**Conclusion for Statement 1:** The statement is true.

### Statement 2:
**Given:** \( \cos x = x^3 + \frac{3}{x} \) for \( x > 0 \) has no solution in \( \mathbb{R} \).

**Step 1:** Recognize that \( \cos x \) ranges from -1 to 1.

**Step 2:** Analyze the expression \( x^3 + \frac{3}{x} \):
- Rewrite it as \( f(x) = x^3 + \frac{3}{x} \).

**Step 3:** Apply the AM-GM inequality:
\[ f(x) = x^3 + 1 + 1 + 1 \geq 4 \sqrt[4]{x^3 \cdot 1 \cdot 1 \cdot 1} = 4. \]

**Step 4:** Since \( f(x) \geq 4 \) and \( \cos x \leq 1 \), it follows that there are no solutions where \( \cos x = f(x) \).

**Conclusion for Statement 2:** The statement is true.

### Statement 3:
**Given:** \( \tan^2 x - \sin^2 x = \tan^2 x \cdot \sin^2 x \).

**Step 1:** Rewrite \( \tan^2 x \):
\[ \tan^2 x = \frac{\sin^2 x}{\cos^2 x}. \]

**Step 2:** Substitute into the equation:
\[ \frac{\sin^2 x}{\cos^2 x} - \sin^2 x = \tan^2 x \cdot \sin^2 x. \]

**Step 3:** Factor out \( \sin^2 x \):
\[ \sin^2 x \left( \frac{1}{\cos^2 x} - 1 \right) = \tan^2 x \cdot \sin^2 x. \]

**Step 4:** Simplify the left side:
\[ \frac{\sin^2 x (1 - \cos^2 x)}{\cos^2 x} = \tan^2 x \cdot \sin^2 x. \]
Since \( 1 - \cos^2 x = \sin^2 x \), we have:
\[ \frac{\sin^4 x}{\cos^2 x} = \tan^2 x \cdot \sin^2 x. \]

**Conclusion for Statement 3:** The statement is true.

### Final Conclusion:
All statements are true.