Question

cotAtan(90-A)-sec(90-A)cosec+√3tan12°tan60°tan78°=2

Answer 1

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$\underline\bold{\huge{ANSWER \: :}}$

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$\boxed{L. H. S.}$

cot A tan (90°-A) - sec (90°-A) cosec A + {√3×tan 12°×tan 60°×tan 78°}

= (cot A × cot A) - (cosec A × cosec A) + {√3 × tan 12° × √3 × tan (90°-12°)}

= cot²A - cosec²A + {3 × tan 12 °× cot 12°}

= - (-cot²A + cosec²A) + {3×1}

= - (cosec²A - cot²A) + 3

= -1 +3

= 3-1

= 2 = $\boxed{R. H. S.}$

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Answer 2

Answer:

L. H. S. =CotATan(90•-A) -sec(90•-A) cosecA+√3Tan12•Tan60•tan78•

=CotA CotA-cosecACosecA+Tan12•√3Tan60•Cot12•

=Cot^A-Cosec^A+Tan12•1/Tan12•√3Tan60•

=Cot^A-Cosec^A+(√3)^

=-1+3

=2. prove