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First Principle of Differentiation

Solve:- 3tan...

Question

Solve:-
$3 tan (θ - 15^{\circ}) = tan (θ + 15^{\circ})$

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Solution

Solution -
$3 t a n (θ - 15^{\circ}) = t a n (θ + 15^{\circ})$

$3 = \frac{t a n (θ + 15^{\circ})}{t a n (θ - 15^{\circ})}$

$\frac{3 + 1}{3 - 1} = \frac{t a n (θ + 15^{\circ}) + t a n (θ - 15^{\circ})}{t a n (θ + 15^{\circ} - t a n (θ - 15^{\circ})}$

$2 = \frac{\frac{s i n (θ + 15^{\circ})}{c o s (θ + 15^{\circ})} + \frac{s i n (θ - 15^{\circ})}{c o s (θ - 15^{\circ})}}{\frac{s i n (θ + 15^{\circ})}{c o s (θ + 15^{\circ})} - \frac{s i n (θ - 15^{\circ})}{c o s (θ - 15^{\circ})}}$

$2 = \frac{s i n (θ + 15^{\circ}) c o s (θ - 15^{\circ}) + s i n (θ - 15^{\circ}) c o s (θ + 15^{\circ})}{s i n (θ + 15^{\circ}) c o s (θ - 15^{\circ}) - s i n (θ - 15^{\circ}) c o s (θ + 15^{\circ})}$

Apply $s i n (A + B) = s i n A c o s B + c o s A s i n B$

$s i n (A - B) = s i n A c o s B - c o s A s i n B$

$2 = \frac{s i n (2 θ)}{s i n (30^{\circ})}$

$s i n 2 θ = 1$

$2 θ = 2 n π + \frac{π}{2}$

$θ = n π + \frac{π}{4}$ in $0 < θ < 90^{\circ} - θ = 45^{\circ}$

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