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Without using trigonometric tables, find the value of the expression: {(cos 70^circ)/(sin 20^circ) + (cos 55^circ cosec 35^circ)/(tan 5^circ tan 25^circ tan 45^circ tan 65^circ tan 85^circ)}

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Question

Without using trigonometric tables, find the value of the expression:

`{(cos 70^circ)/(sin 20^circ) + (cos 55^circ "cosec"  35^circ)/(tan 5^circ tan 25^circ tan 45^circ tan 65^circ tan 85^circ)}`

Sum
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Solution

Given: `{(cos 70^circ)/(sin 20^circ) + (cos 55^circ "cosec"  35^circ)/(tan 5^circ tan 25^circ tan 45^circ tan 65^circ tan 85^circ)}`

Step-wise calculation:

1. cos 70° = sin 20°, so `(cos 70^circ)/(sin 20^circ) = 1`.

2. `cos 55^circ · "cosec" 35^circ = (cos 55^circ)/(sin 35^circ)`. 

But cos 55° = sin 35°, so cos 55° · cosec 35° = 1.

Thus the second term becomes `1/(tan 5^circ tan 25^circ tan 45^circ tan 65^circ tan 85^circ)`.

3. Pairing complementary angles:

tan 5°·tan 85° = 1 and tan 25°·tan 65° = 1 and tan 45° = 1.

Therefore tan 5° tan 25° tan 45° tan 65° tan 85° = 1·1·1 = 1.

4. So the second term = `1/1` = 1.

⇒ 1 + 1 = 2.

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Chapter 12: Trigonometric Ratios of Some Complemantary Angles - EXERCISE 12 [Page 591]

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 12 Trigonometric Ratios of Some Complemantary Angles
EXERCISE 12 | Q 14. | Page 591
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