The value of the following is : `(tan20^(@))^(2)/(("cosec"70^(@))^(2))+((cot20^(@))^(2))/("sec70^(@))^(2)+2tan15^(@).tan45^(@).tan75^(@)`

Options

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

Answer

To solve the given expression:

\[ \frac{(\tan 20^\circ)^2}{(\csc 70^\circ)^2} + \frac{(\cot 20^\circ)^2}{(\sec 70^\circ)^2} + 2 \tan 15^\circ \tan 45^\circ \tan 75^\circ \]

we will follow these steps:

### Step 1: Rewrite the Trigonometric Functions
Using the identities:
- \(\csc \theta = \frac{1}{\sin \theta}\)
- \(\sec \theta = \frac{1}{\cos \theta}\)
- \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)
- \(\cot \theta = \frac{\cos \theta}{\sin \theta}\)

We can rewrite the expression as follows:

\[ \frac{(\tan 20^\circ)^2}{(\csc 70^\circ)^2} = \frac{(\tan 20^\circ)^2}{\left(\frac{1}{\sin 70^\circ}\right)^2} = (\tan 20^\circ)^2 \cdot \sin^2 70^\circ \]

\[ \frac{(\cot 20^\circ)^2}{(\sec 70^\circ)^2} = \frac{(\cot 20^\circ)^2}{\left(\frac{1}{\cos 70^\circ}\right)^2} = (\cot 20^\circ)^2 \cdot \cos^2 70^\circ \]

### Step 2: Simplify the Expression
Now substituting these back into the expression, we have:

\[ (\tan 20^\circ)^2 \cdot \sin^2 70^\circ + (\cot 20^\circ)^2 \cdot \cos^2 70^\circ + 2 \tan 15^\circ \tan 45^\circ \tan 75^\circ \]

Using the identity \(\sin 70^\circ = \cos 20^\circ\) and \(\cos 70^\circ = \sin 20^\circ\), we can rewrite:

\[ (\tan 20^\circ)^2 \cdot \cos^2 20^\circ + (\cot 20^\circ)^2 \cdot \sin^2 20^\circ + 2 \tan 15^\circ \cdot 1 \cdot \tan 75^\circ \]

### Step 3: Further Simplification
Using the identity \(\tan 75^\circ = \cot 15^\circ\):

\[ (\tan 20^\circ)^2 \cdot \cos^2 20^\circ + (\cot 20^\circ)^2 \cdot \sin^2 20^\circ + 2 \tan 15^\circ \cdot \cot 15^\circ \]

Since \(\tan 15^\circ \cdot \cot 15^\circ = 1\):

\[ (\tan 20^\circ)^2 \cdot \cos^2 20^\circ + (\cot 20^\circ)^2 \cdot \sin^2 20^\circ + 2 \]

### Step 4: Use the Pythagorean Identity
Now, we can simplify the first two terms:

\[ (\tan 20^\circ)^2 \cdot \cos^2 20^\circ + \frac{\cos^2 20^\circ}{\sin^2 20^\circ} \cdot \sin^2 20^\circ = \tan^2 20^\circ \cdot \cos^2 20^\circ + \cos^2 20^\circ \]

Factoring out \(\cos^2 20^\circ\):

\[ \cos^2 20^\circ (\tan^2 20^\circ + 1) + 2 \]

Using the identity \(\tan^2 \theta + 1 = \sec^2 \theta\):

\[ \cos^2 20^\circ \cdot \sec^2 20^\circ + 2 = 1 + 2 = 3 \]

### Final Answer
Thus, the value of the expression is:

\[ \boxed{3} \]