CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

First Principle of Differentiation

Solve:- 3tan...

Question

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

Open in App

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 $0 < θ < 90^{\circ} - θ = 45^{\circ}$

Suggest Corrections

thumbs-up

0

Similar questions

3 tan (θ - 15^{o}) = tan (θ + 15^{o})

{tan}^{2} θ - 4 \sqrt{3} tan θ + 3 = 0

tan θ

3 tan (θ - 15^{o}) = tan (θ + 15^{o}), 0 < θ < 90^{o}

θ = 45^{o}

θ = t a n^{- 1} (2 t a n^{2} θ) - t a n^{- 1} (\frac{1}{3} t a n θ)

t a n θ

{tan}^{2} θ + 2 \sqrt{3} tan θ = 1

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Derivative from First Principles

MATHEMATICS

Watch in App

explore_more_icon

Explore more

First Principle of Differentiation

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon