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प्रश्न
Evaluate:
`(tan 70^circ)/(cot 20^circ) + (sin 38^circ)/(cos 52^circ) - 4 sin^2 30^circ`
मूल्यांकन
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उत्तर
Given: `(tan 70^circ)/(cot 20^circ) + (sin 38^circ)/(cos 52^circ) - 4 sin^2 30^circ`.
Step-wise calculation:
1. `(tan 70^circ)/(cot 20^circ) = tan 70^circ · tan 20^circ`
Since `cot 20^circ = 1/(tan 20^circ)`.
But tan 70° = tan (90° – 20°)
= cot 20°
= `1/(tan 20^circ)`
So, tan 70° · tan 20° = 1.
2. `(sin 38^circ)/(cos 52^circ) = (sin 38^circ)/(cos 52^circ)`.
Since cos 52° = sin (90° – 52°)
= sin 38°
This ratio = 1.
3. `4 sin^2 30^circ = 4 xx (1/2)^2`
= `4 xx 1/4`
= 1
Combine:
1 + 1 – 1 = 1
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