how to solve inverse trig equations

In this post ,we will check what method or ways can be used to solve inverse trig equations

Method 1

1. Obtain the equation in the form given above using inverse trigonometric function identities
i.e
$sin^{-1} x =”value”$
$cos^{-1} x=”value”$
$tan^{-1} x=”value”$
$cosec^{-1} x=”value”$
$sec^{-1} x=”value”$
$cot^{-1} x=”value”$

2. Now we can easily find the value of x from inverse trigonometric function domain and range

Example

$(tan^{-1} x)^2 + (cot^{-1} x)^2 = \frac {5\pi^2}{8}$

Solution

$(tan^{-1} x)^2 + (cot^{-1} x)^2 = \frac {5\pi^2}{8}$
$(tan^{-1} x)^2 + (\frac {\pi}{2} – tan^{-1} x)^2 =\frac {5\pi^2}{8}$
$2tan^{-1} x)^2 – \pi tan^{-1} x -\frac {3\pi ^2}{8}=0$
So $tan^{-1} x= – \pi/4 $ or $tan^{-1} x= 3\pi/4 $
Only possible value is
$tan^{-1} x= – \pi/4 $
So x=-1

Method 2

Sometimes , we can convert the inverse trigonometric equation into algebraic equation by taking, sin ,cos , tan on the both the side and apply suitable trigonometric identities
Then solve the algebraic equation for values of x
Validate the solution

Example

$tan^{-1} (x+1 ) + tan^{-1} (x-1 )= tan^{-1} \frac {8}{31}$

Solution

Taking tan on both the side and applying the trigonometric identity
$tan (A +B) = \frac { tan A + tan B}{1- tan A tan B}$

$\frac {x+1 + x-1}{1- (x+1)(x-1)} = \frac {8}{31}$
or
$\frac {x+1 + x-1}{1- (x+1)(x-1)} = \frac {8}{31}$
Solving this, we get
x=-8 or 1/4

After validating the solution , we found 1/4 as the solution only