$tan 20^{0} tan 40^{0} tan 80^{0} = tan x^{0}$ . Find $x$

Open in App

Solution

$tan 20^{0} tan 40^{0} tan 80^{0}$
$= tan 20^{0} tan (60^{0} - 20^{0}) tan (60^{0} + 20^{0}) = tan (3 \times 20^{0}) = tan 60^{0}$
Ans: 60

Suggest Corrections

Similar questions

(sec x^{0} - tan x^{0}) (sec x^{0} + tan x^{0})

{cos}^{2} x \frac{d y}{d x} + y = tan x (0 \leq x < \frac{π}{2})

{cos}^{2} x \frac{d y}{d x} + y = tan x (0 \leq x \leq \frac{π}{2})

\frac{d y}{d x} + (sec x) y = tan x (0 \leq x \leq \frac{π}{2})

For each of the given differential equation find the general solution.
$\frac{d y}{d x} + y s e c x = t a n x (0 \leq x \leq \frac{π}{2}) .$

Join BYJU'S Learning Program