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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)}`
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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.
