Question

If $y=\tan x.tan2x.tan3x$, then $\frac{dy}{dx}$ has the value equal to

• A. $6se{c}^{2}3x.\tan x.\tan 2x+2{\sec }^{2}x.\tan 2x.\tan 3x+4{\sec }^{2}2x.\tan 3x.\tan x$
• B. $2y(cosec2x+2cosec4x+3cosec6x)$
• C. $3se{c}^{2}3x-2{\sec }^{2}2x-{\sec }^{2}x$
• D. ${\sec }^{2}x+2{\sec }^{2}2x+3{\sec }^{2}3x$

Answer 1

The correct option is C $3se{c}^{2}3x-2{\sec }^{2}2x-{\sec }^{2}x$

$y=\tan x\tan 2x\tan 3x$
We know
$\tan 3x=\tan (x+2x)$
$=\frac{\tan x+\tan 2x}{1-\tan x.\tan 2x}$
By cross multiplication, we get
$\tan 3x-\tan x\tan 2x\tan 3x=\tan x+\tan 2x$
$\tan x\tan 2x\tan 3x=\tan 3x-\tan 2x-\tan x=y$
$\frac{dy}{dx}=\frac{d}{dx}\left(\tan 3x-\tan 2x-\tan x\right)$
$=3{\sec }^{2}3x-2{\sec }^{2}2x-{\sec }^{2}x$ Option (c)