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प्रश्न
Evaluate: `cos 58^@/sin 32^@ + sin 22^@/cos 68^@ - (cos 38^@ cosec 52^@)/(tan 18^@ tan 35^@ tan 60^@ tan 72^@ tan 65^@)`
उत्तर
cos 58° = cos (90° - 32°) = sin 32°
sin 22° = sin (90° - 68°) = cos 68°
cos 38° = cos (90 – 52) = sin 52°
tan 18° = cot 72 tan 35° = cot 55°
`=> sin 32^@/sin 32^@ + cos 68^@/cos 68^@ - (sin 52 cosec 52)/(tan 72.cot72 tan 55 cot 55.tan 60)`
`= 1 + 1 - 1/sqrt3 = (2sqrt3-1)/sqrt3 xx sqrt3/sqrt3 = (6 - sqrt3)/3`
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