मराठी

Without using trigonometric tables, find the value of the expression: {(cos 65^circ)/(sin 25^circ) + (cosec 34^circ)/(sec 56^circ) – (2 cos 43^circ cosec 47^circ)

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प्रश्न

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

`{(cos 65^circ)/(sin 25^circ) + ("cosec"  34^circ)/(sec 56^circ) - (2 cos 43^circ "cosec"  47^circ)/(tan 10^circ tan 40^circ tan 50^circ tan 80^circ)}`

बेरीज
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उत्तर

Given: `{(cos 65^circ)/(sin 25^circ) + ("cosec"  34^circ)/(sec 56^circ) - (2 cos 43^circ "cosec"  47^circ)/(tan 10^circ tan 40^circ tan 50^circ tan 80^circ)}`

Step-wise calculation:

1. `(cos 65^circ)/(sin 25^circ) = (cos 65^circ)/(cos 65^circ) = 1`   ...(Since sin 25° = cos 65°)

2. `("cosec"  34^circ)/(sec 56^circ) = (1/sin 34^circ)/(1/cos 56^circ)`

= `(cos 56^circ)/(sin 34^circ) = 1`   ...(Since cos 56° = sin 34°)

3. For the third term numerator:

`2 cos 43^circ · "cosec"  47^circ = 2 · cos 43^circ · (1/sin 47^circ)` 

= `2 · (cos 43^circ/sin 47^circ)` 

= 2 · 1

= 2 because sin 47° = cos 43°.

4. Denominator: tan 10° · tan 40° · tan 50° · tan 80°

= (tan 10° · tan 80°) · (tan 40° · tan 50°)

= (tan 10° · cot 10°) · (tan 40° · cot 40°)

= 1 · 1

= 1

5. Hence the third fraction = `2/1` = 2.

Combine terms: 1 + 1 – 2 = 0.

The value of the expression is 0.

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पाठ 12: Trigonometric Ratios of Some Complemantary Angles - EXERCISE 12 [पृष्ठ ५९१]

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आर. एस. अग्रवाल Mathematics [English] Class 10
पाठ 12 Trigonometric Ratios of Some Complemantary Angles
EXERCISE 12 | Q 16. | पृष्ठ ५९१
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