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प्रश्न
Prove that:
`cot theta tan (90^\circ - theta) - sec (90^\circ - theta) "cosec" theta + sqrt 3 tan 12^\circ tan 60^\circ tan 78^circ` = 2
योग
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उत्तर
LHS = `cot theta tan (90^\circ - theta) - sec (90^\circ - theta) "cosec" theta + sqrt 3 tan 12^\circ tan 60^\circ tan 78^circ`
= `cot theta * cot theta - "cosec" theta * "cosec" theta + sqrt 3 tan 12^\circ * sqrt 3 tan 78^\circ`
= `cot^2 theta - "cosec'^2 theta + 3 tan 12^\circ tan 78^\circ`
= `cot^2 theta - (1 + cot^2 theta) + 3 tan 12^\circ cot 12^\circ`
= −1 + 3(1)
= 2
= RHS
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