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 \(\boxed{1}\).
Statement-I: In the interval \( [0, 2\pi] \), the number of common solutions of the equations
\[ 2\sin^2\theta - \cos 2\theta = 0 \]
and
\[ 2\cos^2\theta - 3\sin\theta = 0 \]
is two.
Statement-II: The number of solutions of
\[ 2\cos^2\theta - 3\sin\theta = 0 \]
in \( [0, \pi] \) is two.
If \( A \) and \( B \) are acute angles satisfying
\[ 3\cos^2 A + 2\cos^2 B = 4 \]
and
\[ \frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A}, \]
Then \( A + 2B = \ ? \)
If
\[ \sin \theta + 2 \cos \theta = 1 \]
and
\[ \theta \text{ lies in the 4\textsuperscript{th} quadrant (not on coordinate axes), then } 7 \cos \theta + 6 \sin \theta =\ ? \]