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Find the value of tan 3θ if sec 3θ= cosec (4θ-15°). |
$\frac{1}{\sqrt{3}}$ $\sqrt{3}$ -1 1 |
1 |
Given :- sec 3θ= cosec (4θ-15°) If secA = cosecB then A + B = 90º So , 3θ + 4θ - 15° = 90° 7θ = 105° θ = 15° Now , tan 3θ = tan 45° = 1 |
