Concept:

Pythagorous Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. 

In triangle ABC with sides a, b, c, the sine rule:

\(\rm \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k\)

sin A - cos B =\(\rm 2\cos(\frac{A +B}{2})\sin(\frac{A -B}{2})\)

Calculation:

1. We have,

sin \(\rm B = \frac 1 3=\frac{P}{H}\)

Therefore, the right-angled triangle can be drawn as follows,

Here, P = 1
H = 3
B = x

Using the Pythagorean theorem,

3² = 1² + x²

x² = 8

\(\therefore x=2\sqrt{2}\)

Now , Cosec C = \(\frac{3}{2\sqrt{2}}\)

Therefore, statement (1) is not correct.

2. It is given that,

b cos B = c cos C

Using the sine rule,

\(\rm \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{K}\)

⇒ b = K sin B and c = K sin C

⇒ 2 sin B cos B = 2 sin C cos C

Let,
K = 2

⇒ 2 sin B cos B = 2 sin C cos C

⇒ sin 2B = sin 2C  [2 sin x cos x = sin 2x]

⇒ sin 2B - sin 2C = 0

Using the formula:

sin C - sin D = 2 cos\((\frac{C+D}{2})\)× sin\((\frac{C-D}{2})\)

⇒ 2 cos (B + C) sin (B - C) = 0

In this case,

either,

cos (B + C) = 0

Therefore, (B + C) = 90° .... (1)

or,

or sin (B -C) = 0

B - C = 0

⇒ B = C  .... (2)

From equations (1) and (2),

B = C = 45°

Therefore, ABC must be isosceles

Thus, option (2) is correct