2 arctan(x) = arctan($\frac{2x}{1 - x^{2}}$) = arcsin($\frac{2x}{1 + x^{2}}$) = arccos($\frac{1 - x^{2}}{1 + x^{2}}$)

We will learn how to prove the property of the inverse trigonometric function, 2 arctan(x) = arctan($\frac{2x}{1 - x^{2}}$) = arcsin($\frac{2x}{1 + x^{2}}$) = arccos($\frac{1 - x^{2}}{1 + x^{2}}$)

or, 2 tan$^{-1}$ x = tan$^{-1}$ ($\frac{2x}{1 - x^{2}}$) = sin$^{-1}$ ($\frac{2x}{1 + x^{2}}$) = cos$^{-1}$ ($\frac{1 - x^{2}}{1 + x^{2}}$)

Proof: 

Let, tan$^{-1}$ x = θ         

Therefore, tan θ = x

We know that,

tan 2θ = $\frac{2 tan θ}{1 - tan^{2}θ}$

tan 2θ = $\frac{2x}{1 - x^{2}}$

2θ = tan$^{-1}$($\frac{2x}{1 - x^{2}}$)

2 tan$^{-1}$ x = tan$^{-1}$($\frac{2x}{1 - x^{2}}$) …………………….. (i)

Again, sin 2θ = $\frac{2 tan θ}{1 + tan^{2}θ}$

sin 2θ = $\frac{2x}{1 + x^{2}}$

2θ = sin$^{-1}$($\frac{2x}{1 + x^{2}}$ )

2 tan$^{-1}$ x = sin$^{-1}$($\frac{2x}{1 + x^{2}}$) …………………….. (ii)

Now,  cos 2θ = $\frac{1 - tan^{2}θ}{1 + tan^{2}θ}$

 cos 2θ =  $\frac{1 - x^{2} }{1 + x^{2} }$

2θ = cos$^{-1}$ ($\frac{1 - x^{2} }{1 + x^{2} }$)

2 tan$^{-1}$ x = cos ($\frac{1 - x^{2} }{1 + x^{2} }$) …………………….. (iii)

Therefore, from (i), (ii) and (iii) we get, 2 tan$^{-1}$ x = tan$^{-1}$ $\frac{2x}{1 - x^{2}}$ = sin$^{-1}$ $\frac{2x}{1 + x^{2}}$ = cos$^{-1}$ $\frac{1 - x^{2}}{1 + x^{2}}$                   Proved.

Solved examples on property of inverse circular function 2 arctan(x) = arctan($\frac{2x}{1 - x^{2}}$) = arcsin($\frac{2x}{1 + x^{2}}$) = arccos($\frac{1 - x^{2}}{1 + x^{2}}$):

1. Find the value of the inverse function tan(2 tan$^{-1}$ $\frac{1}{5}$).

Solution:

tan (2 tan$^{-1}$ $\frac{1}{5}$)

= tan (tan$^{-1}$ $\frac{2 × \frac{1}{5}}{1 - (\frac{1}{5})^{2}}$), [Since, we know that, 2 tan$^{-1}$ x = tan$^{-1}$($\frac{2x}{1 - x^{2}}$)]

 = tan (tan$^{-1}$ $\frac{\frac{2}{5}}{1 - \frac{1}{25}}$)

= tan (tan$^{-1}$ $\frac{5}{12}$)

= $\frac{5}{12}$

2. Prove that, 4 tan$^{-1}$ $\frac{1}{5}$ - tan$^{-1}$ $\frac{1}{70}$ + tan$^{-1}$ $\frac{1}{99}$ = $\frac{π}{4}$

Solution:

L. H. S. = 4 tan$^{-1}$ $\frac{1}{5}$ - tan$^{-1}$ $\frac{1}{70}$ + tan$^{-1}$ $\frac{1}{99}$

= 2(2 tan$^{-1}$ $\frac{1}{5}$) - tan$^{-1}$ $\frac{1}{70}$ + tan$^{-1}$ $\frac{1}{99}$

= 2(tan$^{-1}$ $\frac{2 × \frac{1}{5}}{1 - (\frac{1}{5})^{2}}$) - tan$^{-1}$ $\frac{1}{70}$ + tan$^{-1}$ $\frac{1}{99}$, [Since, 2 tan$^{-1}$ x = tan$^{-1}$($\frac{2x}{1 - x^{2}}$)]

= 2 (tan$^{-1}$ $\frac{2\frac{1}{5}}{1 - (\frac{1}{25})}$)- tan$^{-1}$ $\frac{1}{70}$ + tan$^{-1}$ $\frac{1}{99}$,

= 2 tan$^{-1}$ $\frac{5}{12}$ - (tan$^{-1}$ $\frac{1}{70}$ - tan$^{-1}$ $\frac{1}{99}$)

= tan$^{-1}$ ($\frac{2 × \frac{5}{12}}{1 - (\frac{5}{12})^{2}}$) - tan$^{-1}$ ($\frac{\frac{1}{70} - \frac{1}{99}}{1 + \frac{1}{77} × \frac{1}{99}}$)

= tan$^{-1}$ $\frac{120}{199}$ - tan$^{-1}$ $\frac{29}{6931}$

= tan$^{-1}$ $\frac{120}{199}$ - tan$^{-1}$ $\frac{1}{239}$

= tan$^{-1}$ ($\frac{\frac{120}{199} - \frac{1}{239}}{1 + \frac{120}{119} × \frac{1}{239}}$)

= tan$^{-1}$ 1

= tan$^{-1}$ (tan $\frac{π}{4}$)

= $\frac{π}{4}$ = R. H. S.              Proved.

● Inverse Trigonometric Functions

• General and Principal Values of sin$^{-1}$ x
• General and Principal Values of cos$^{-1}$ x
• General and Principal Values of tan$^{-1}$ x
• General and Principal Values of csc$^{-1}$ x
• General and Principal Values of sec$^{-1}$ x
• General and Principal Values of cot$^{-1}$ x
• Principal Values of Inverse Trigonometric Functions
• General Values of Inverse Trigonometric Functions
  
• arcsin(x) + arccos(x) = $\frac{π}{2}$
• arctan(x) + arccot(x) = $\frac{π}{2}$
• arctan(x) + arctan(y) = arctan($\frac{x + y}{1 - xy}$)
• arctan(x) - arctan(y) = arctan($\frac{x - y}{1 + xy}$)
• arctan(x) + arctan(y) + arctan(z)= arctan$\frac{x + y + z – xyz}{1 – xy – yz – zx}$
• arccot(x) + arccot(y) = arccot($\frac{xy - 1}{y + x}$)
• arccot(x) - arccot(y) = arccot($\frac{xy + 1}{y - x}$)
• arcsin(x) + arcsin(y) = arcsin(x $\sqrt{1 - y^{2}}$ + y$\sqrt{1 - x^{2}}$)
• arcsin (x) - arcsin(y) = arcsin (x $\sqrt{1 - y^{2}}$ - y$\sqrt{1 - x^{2}}$)
• arccos (x) + arccos(y) = arccos(xy - $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$)
• arccos(x) - arccos(y) = arccos(xy + $\sqrt{1 - x^{2}}$$\sqrt{1 - y^{2}}$)
• 2 arcsin(x) = arcsin(2x$\sqrt{1 - x^{2}}$) 
• 2 arccos(x) = arccos(2x$^{2}$ - 1)
• 2 arctan(x) = arctan($\frac{2x}{1 - x^{2}}$) = arcsin($\frac{2x}{1 + x^{2}}$) = arccos($\frac{1 - x^{2}}{1 + x^{2}}$)
• 3 arcsin(x) = arcsin(3x - 4x$^{3}$)
• 3 arccos(x) = arccos(4x$^{3}$ - 3x)
• 3 arctan(x) = arctan($\frac{3x - x^{3}}{1 - 3 x^{2}}$)
• Inverse Trigonometric Function Formula
• Principal Values of Inverse Trigonometric Functions
• Problems on Inverse Trigonometric Function

11 and 12 Grade Math

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