Question

without using Trigonometric Table evaluate sec 19 / Cos Sector 17 + 1/Root 3 × tan 39 × tan 25 × tan 30 × sin 65×tan 51 - 3 ( sin square 41 + sin square 49)

Answer 1

Hi,

Sin 18°/cos 72° + √3 [tan10 tan 30 tan 40 tan 50 tan 80]

=Sin 18°/cos(90-72) + √3 [tan10 tan 30 tan 40 tan(90-50) tan(90-80)]

=Sin 18°/Sin 18 + √3 [tan10 tan 30 tan 40 cot40 cot10]

=1 + √3 [tan10 tan 30 tan 40 1/tan40 1/tan10]

=1 + √3 [tan 30]

=1 + √3 [1/√3]

=1 + 1 = 2

mark as brainliest!!

Answer 2

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