Question

Write the simplest form of tan inverse of 3cosx - 4 sinx/ 4 cosx +3sinx

Answer 1

Answer:

Step-by-step explanation:

Answer 2

∴ the simplest from is $tan^{-\frac{3}{4} } -x$

Step-by-step explanation:

Given;

$tan^{-\frac{3cosx-4sinx}{4cosx+3sinx} }=?$

Let;

   $a=\frac{3cosx-4sinx}{4cosx+3sinx}$

⇒$a=\frac{3-4tanx}{4+3tanx}$  (∵ By dividing numerator and denominator with $cosx$)

⇒ $a=\frac{\frac{3}{4} -tanx}{1+\frac{3}{4} tanx}$  (∵ By dividing numerator and denominator with $4$)

⇒$a=\frac{tany -tanx}{1+tany tanx}$  Where $tany=\frac{3}{4}$

⇒$a=tan(y-x)$   (∵$\frac{tana -tanb}{1+tana tanb}=tan(a-b)$ )

From main equation;

$tan^{-\frac{3cosx-4sinx}{4cosx+3sinx} }=tan^{-a}$

Plug the value of $a$;

$=tan^{-tan(y-x)}$

$=(y-x)$  since range of $tan$ values lies between $(-\frac{\pi}{2} ,\frac{\pi}{2} )$

$=tan^{-\frac{3}{4} } -x$  (∵$y=tan^{-\frac{3}{4} }$ )

So the simplest from is $tan^{-\frac{3}{4} } -x$