What is the value of `(tan^(2)25^(@))/(cosec^(2)65^(@))+(cot^(2)25^(@))/(sec^(2)65^(@))+tan20^(@)45^(@)tan70^(@)` ?

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the expression

\[ \frac{\tan^2 25^\circ}{\csc^2 65^\circ} + \frac{\cot^2 25^\circ}{\sec^2 65^\circ} + \tan 20^\circ \tan 45^\circ \tan 70^\circ, \]

we will break it down step by step.

### Step 1: Simplify the first term
We know that \(\csc^2 65^\circ = \frac{1}{\sin^2 65^\circ}\). Since \(65^\circ\) is complementary to \(25^\circ\) (i.e., \(25^\circ + 65^\circ = 90^\circ\)), we can use the identity \(\sin 65^\circ = \cos 25^\circ\). Thus,

\[ \csc^2 65^\circ = \frac{1}{\sin^2 65^\circ} = \frac{1}{\cos^2 25^\circ}. \]

Now substituting this into the first term:

\[ \frac{\tan^2 25^\circ}{\csc^2 65^\circ} = \tan^2 25^\circ \cdot \cos^2 25^\circ. \]

Using the identity \(\tan^2 x = \frac{\sin^2 x}{\cos^2 x}\), we have:

\[ \tan^2 25^\circ = \frac{\sin^2 25^\circ}{\cos^2 25^\circ}. \]

Thus,

\[ \frac{\tan^2 25^\circ}{\csc^2 65^\circ} = \frac{\sin^2 25^\circ}{\cos^2 25^\circ} \cdot \cos^2 25^\circ = \sin^2 25^\circ. \]

### Step 2: Simplify the second term
For the second term, we know that \(\sec^2 65^\circ = \frac{1}{\cos^2 65^\circ}\). Again, using the complementary angle identity, \(\cos 65^\circ = \sin 25^\circ\), we have:

\[ \sec^2 65^\circ = \frac{1}{\sin^2 25^\circ}. \]

Thus,

\[ \frac{\cot^2 25^\circ}{\sec^2 65^\circ} = \cot^2 25^\circ \cdot \sin^2 25^\circ. \]

Using the identity \(\cot^2 x = \frac{\cos^2 x}{\sin^2 x}\):

\[ \cot^2 25^\circ = \frac{\cos^2 25^\circ}{\sin^2 25^\circ}. \]

So we have:

\[ \frac{\cot^2 25^\circ}{\sec^2 65^\circ} = \frac{\cos^2 25^\circ}{\sin^2 25^\circ} \cdot \sin^2 25^\circ = \cos^2 25^\circ. \]

### Step 3: Simplify the third term
Now, we simplify the third term:

\[ \tan 20^\circ \tan 45^\circ \tan 70^\circ. \]

Since \(\tan 45^\circ = 1\), we have:

\[ \tan 20^\circ \tan 45^\circ \tan 70^\circ = \tan 20^\circ \cdot 1 \cdot \tan 70^\circ = \tan 20^\circ \tan 70^\circ. \]

Using the identity \(\tan(90^\circ - x) = \cot x\), we have \(\tan 70^\circ = \cot 20^\circ\). Thus:

\[ \tan 20^\circ \tan 70^\circ = \tan 20^\circ \cdot \cot 20^\circ = 1. \]

### Step 4: Combine all terms
Now we can combine all the simplified terms:

\[ \sin^2 25^\circ + \cos^2 25^\circ + 1. \]

Using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\):

\[ \sin^2 25^\circ + \cos^2 25^\circ = 1. \]

Thus, we have:

\[ 1 + 1 = 2. \]

### Final Answer
The value of the expression is

\[ \boxed{2}. \]