To evaluate the expression \( (\sin 70^\circ)(\cot 10^\circ \cot 70^\circ - 1) \), we proceed with the following steps:
1. Simplifying the Cotangent Product:
We use the trigonometric identity for the product of cotangents:
\[ \cot A \cot B - 1 = \frac{\cos A \cos B}{\sin A \sin B} - 1 = \frac{\cos A \cos B - \sin A \sin B}{\sin A \sin B} = \frac{\cos(A+B)}{\sin A \sin B} \]
Applying this to our expression with \( A = 10^\circ \) and \( B = 70^\circ \):
\[ \cot 10^\circ \cot 70^\circ - 1 = \frac{\cos(10^\circ + 70^\circ)}{\sin 10^\circ \sin 70^\circ} = \frac{\cos 80^\circ}{\sin 10^\circ \sin 70^\circ} \]
2. Substituting Back into the Original Expression:
Now, multiply by \( \sin 70^\circ \):
\[ (\sin 70^\circ)\left( \frac{\cos 80^\circ}{\sin 10^\circ \sin 70^\circ} \right) = \frac{\cos 80^\circ}{\sin 10^\circ} \]
3. Using Complementary Angle Identity:
We know that \( \cos 80^\circ = \sin(90^\circ - 80^\circ) = \sin 10^\circ \):
\[ \frac{\cos 80^\circ}{\sin 10^\circ} = \frac{\sin 10^\circ}{\sin 10^\circ} = 1 \]
Final Answer:
The value of the expression is \(\boxed{1}\).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,