Question

Prove that tan $\text{tan}4\theta =\frac{4\text{tan}\theta (1-{\text{tan}}^2\theta )}{1-6{\text{tan}}^2\theta +{\text{tan}}^4\theta }$. Find the angle $\theta ϵ(0,\frac{\pi }{2})$, in which the given proof does not hold.

Or

Prove that $\frac{\text{cos}4x+\text{cos}3x+\text{cos}2x}{\text{sin}4x+\text{sin}3x+\text{sin}2x}$ = cot 3x . Do you think at $x=\frac{\pi }{3}$, the given proof holds true?

Answer 1

We have, LHS = tan 4 $\theta$ = tan 2 (2 $\theta$) = $\frac{2\text{tan}2\theta }{1-{\text{tan}}^{2}2\theta }[\because \text{tan}2\theta =\frac{2\text{tan}\theta }{1-{\text{tan}}^2\theta }]$

= $\frac{2\left(\frac{2tan\theta }{1-ta{n}^2\theta }\right)}{1-{\left(\frac{2tan\theta }{1-ta{n}^2\theta }\right)}^2}=\frac{\frac{4tan\theta }{1-ta{n}^2\theta }}{\frac{(1-ta{n}^2\theta {)}^2-4ta{n}^2\theta }{(1-ta{n}^2\theta {)}^2}}$

= $\frac{4tan\theta (1-ta{n}^2\theta )}{(1-ta{n}^2\theta {)}^2-4ta{n}^2\theta }=\frac{4tan\theta (1-ta{n}^2\theta )}{1-6ta{n}^2\theta +ta{n}^4\theta }=\text{RHS}\text{Hence proved.}$

We know that tan $\theta$ is not defined at $\theta$ = $\frac{\pi }{2}$

$\therefore$ tan 4 $\theta$ is not defined at $\theta$ = $\frac{\pi }{8}ϵ(0,\frac{\pi }{2})$

Hence, given proof does not exist at $\theta$ = $\frac{\pi }{8}$

Or

We have,

LHS = $\frac{cos4x+cos3x+cos2x}{sin4x+sin3x+sin2x}$

$=\frac{cos4x+cos2x+cos3x}{sin4x+sin2x+sin3x}$

$=\frac{2cos\left(\frac{4x+2x}{2}\right)cos\left(\frac{4x-2x}{2}\right)+cos3x}{2sin\left(\frac{4x+2x}{2}\right)cos\left(\frac{4x-2x}{2}\right)+sin3x}$

$[\because cosC+cosD=2cos\left(\frac{C+D}{2}\right).cos\left(\frac{C-D}{2}\right)\text{and}sinC+sinD=2sin(\frac{C+D}{2}.).cos\left(\frac{C-D}{2}\right)]$

= $\frac{2cos3xcosx+cos3x}{2sin3xcosx+sin3x}$

= $\frac{cos3x(2cosx+1)}{sin3x(2cosx+1)}=\frac{cos3x}{sin3x}=cot3x=\text{RHS}$

At $x=\frac{\pi }{3},\text{cot 3}\times \frac{\pi }{3}=\text{cot}\pi =\infty .$ Hence, at $x=\frac{\pi }{3}$ given proof does not hold true.