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Question
Find the value of `(cos 38° cosec 52°)/(tan 18° tan 35° tan 60° tan 72° tan 55°)`
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Solution
`(cos 38° cosec 52°)/(tan 18° tan 35° tan 60° tan 72° tan 55°)`
`= ( cos 38 ° sec (90°-52°))/( cot (90° -18° ) cot (90° -35° ) tan 60° tan 72° tan 55°)`
=` (cos 38° sec 38°)/( cot 72° cot 55° tan 60° tan 72° tan 55°)`
=`(cos 38° xx1/(cos 38°))/(1/(tan 72°) xx1/( tan 55°) xx sqrt(3 ) xx tan 72° xx tan 55°)`
=`1/sqrt(3)`
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
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But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
