Without using trigonometric tables , evaluate :
(i) `(sin40^(@))/(cos50^(@))+(3tan50^(@))/(cot40^(@))`
(ii) `(sec37^(@))/("cosec "53^(@))-(tan20^(@))/(cot70^(@))`
(iii) `(cos74^(@))/(sin16^(@))+(sin12^(@))/(cos78^(@))-sin18^(@)sec72^(@)`
(iv) `sin35^(@)sec55^(@)+(4sec32^(@))/("cosec "58^(@))`

Answer

Let's evaluate the given expressions step by step without using trigonometric tables.

### (i) Evaluate:
\[ \frac{\sin 40^\circ}{\cos 50^\circ} + \frac{3 \tan 50^\circ}{\cot 40^\circ} \]

**Step 1:** Rewrite \(\cos 50^\circ\) using the co-function identity:
\[ \cos 50^\circ = \sin(90^\circ - 50^\circ) = \sin 40^\circ \]
Thus, we can rewrite the first term:
\[ \frac{\sin 40^\circ}{\cos 50^\circ} = \frac{\sin 40^\circ}{\sin 40^\circ} = 1 \]

**Step 2:** Rewrite \(\tan 50^\circ\) and \(\cot 40^\circ\):
\[ \tan 50^\circ = \frac{\sin 50^\circ}{\cos 50^\circ} \quad \text{and} \quad \cot 40^\circ = \frac{1}{\tan 40^\circ} = \frac{\cos 40^\circ}{\sin 40^\circ} \]
Thus:
\[ \frac{3 \tan 50^\circ}{\cot 40^\circ} = 3 \tan 50^\circ \cdot \tan 40^\circ = 3 \cdot \frac{\sin 50^\circ}{\cos 50^\circ} \cdot \frac{\sin 40^\circ}{\cos 40^\circ} \]

**Step 3:** Using the identity \(\tan(90^\circ - \theta) = \cot \theta\):
\[ \tan 50^\circ = \cot 40^\circ \Rightarrow \tan 50^\circ \cdot \cot 40^\circ = 1 \]
Thus:
\[ \frac{3 \tan 50^\circ}{\cot 40^\circ} = 3 \]

**Step 4:** Combine the results:
\[ 1 + 3 = 4 \]

### Final Answer for (i):
\[ 4 \]

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### (ii) Evaluate:
\[ \frac{\sec 37^\circ}{\csc 53^\circ} - \frac{\tan 20^\circ}{\cot 70^\circ} \]

**Step 1:** Rewrite \(\csc 53^\circ\):
\[ \csc 53^\circ = \frac{1}{\sin 53^\circ} = \frac{1}{\cos(90^\circ - 53^\circ)} = \frac{1}{\sin 37^\circ} \]
Thus:
\[ \frac{\sec 37^\circ}{\csc 53^\circ} = \sec 37^\circ \cdot \sin 37^\circ = \frac{1}{\cos 37^\circ} \cdot \sin 37^\circ = \tan 37^\circ \]

**Step 2:** Rewrite \(\tan 20^\circ\) and \(\cot 70^\circ\):
\[ \cot 70^\circ = \tan(90^\circ - 70^\circ) = \tan 20^\circ \]
Thus:
\[ \frac{\tan 20^\circ}{\cot 70^\circ} = \frac{\tan 20^\circ}{\tan 20^\circ} = 1 \]

**Step 3:** Combine the results:
\[ \tan 37^\circ - 1 \]

### Final Answer for (ii):
\[ 0 \]

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### (iii) Evaluate:
\[ \frac{\cos 74^\circ}{\sin 16^\circ} + \frac{\sin 12^\circ}{\cos 78^\circ} - \sin 18^\circ \sec 72^\circ \]

**Step 1:** Rewrite \(\cos 74^\circ\) and \(\sin 16^\circ\):
\[ \cos 74^\circ = \sin(90^\circ - 74^\circ) = \sin 16^\circ \]
Thus:
\[ \frac{\cos 74^\circ}{\sin 16^\circ} = \frac{\sin 16^\circ}{\sin 16^\circ} = 1 \]

**Step 2:** Rewrite \(\cos 78^\circ\):
\[ \cos 78^\circ = \sin(90^\circ - 78^\circ) = \sin 12^\circ \]
Thus:
\[ \frac{\sin 12^\circ}{\cos 78^\circ} = \frac{\sin 12^\circ}{\sin 12^\circ} = 1 \]

**Step 3:** Rewrite \(\sec 72^\circ\):
\[ \sec 72^\circ = \frac{1}{\cos 72^\circ} = \frac{1}{\sin(90^\circ - 72^\circ)} = \frac{1}{\sin 18^\circ} \]
Thus:
\[ \sin 18^\circ \sec 72^\circ = \sin 18^\circ \cdot \frac{1}{\sin 18^\circ} = 1 \]

**Step 4:** Combine the results:
\[ 1 + 1 - 1 = 1 \]

### Final Answer for (iii):
\[ 1 \]

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### (iv) Evaluate:
\[ \sin 35^\circ \sec 55^\circ + \frac{4 \sec 32^\circ}{\csc 58^\circ} \]

**Step 1:** Rewrite \(\sec 55^\circ\):
\[ \sec 55^\circ = \frac{1}{\cos 55^\circ} = \frac{1}{\sin(90^\circ - 55^\circ)} = \frac{1}{\sin 35^\circ} \]
Thus:
\[ \sin 35^\circ \sec 55^\circ = \sin 35^\circ \cdot \frac{1}{\sin 35^\circ} = 1 \]

**Step 2:** Rewrite \(\csc 58^\circ\):
\[ \csc 58^\circ = \frac{1}{\sin 58^\circ} = \frac{1}{\cos(90^\circ - 58^\circ)} = \frac{1}{\cos 32^\circ} \]
Thus:
\[ \frac{4 \sec 32^\circ}{\csc 58^\circ} = 4 \cdot \frac{1}{\cos 32^\circ} \cdot \sin 58^\circ = 4 \]

**Step 3:** Combine the results:
\[ 1 + 4 = 5 \]

### Final Answer for (iv):
\[ 5 \]

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### Summary of Answers:
1. \(4\)
2. \(0\)
3. \(1\)
4. \(5\)

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