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 \(\boxed{1}\).
Given that $\sin \theta + \cos \theta = x$, prove that $\sin^4 \theta + \cos^4 \theta = \dfrac{2 - (x^2 - 1)^2}{2}$.
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
The term independent of $ x $ in the expansion of $$ \left( \frac{x + 1}{x^{3/2} + 1 - \sqrt{x}} \cdot \frac{x + 1}{x - \sqrt{x}} \right)^{10} $$ for $ x>1 $ is: