Question

tan(90-∅).cot∅-sec(90-∅).cosec∅+
$\sqrt{3} \tan(12) \times \tan(60) \times \tan(78)$
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Answer 1

tan(90-∅).cot∅-sec(90-∅).cosec∅+

$\sqrt{3} \tan(12) \times \tan(60) \times \tan(78)$

=cot∅cot∅-cosec∅cosec∅+√3tan(90-78)tan78tan60°

= cot²∅ - cosec²∅+√3cot78tan78(√3)

= -1 + √3×√3
= -1 + 3

[ Since , cot²A- cosec²A = -1]

= 2

••••

Answer 2

Answer:

2

Step-by-step explanation:

tan (90-∅). cot∅ - sec (90-∅). cosec∅ + √3 * tan(12) * tan(60) * tan(78)

= cot∅ . cot∅ - cosec∅ . cosec∅ + √3 * tan(90-78) * tan(60) * tan(78)

[as tan (90 - ∅) = cot∅ and sec (90-∅) = cosec∅]

= cot²∅ - cosec²∅ + √3 * cot(78) * tan(78) * tan(60)

= -1 + √3 * 1 * tan(60)

[as cot²∅ - cosec²∅ = 1 and cot∅ * tan∅ ]

= -1 + √3 * √3 [as tan(60) = √3 ]

= -1 + 3

= 2