# Inverse Tips and Tricks and Shortcuts

## Tips and Tricks for Inverse

Here, In this Page you will get to know about the Tips, Tricks and shortcuts for for Inverse . Also you will learn about the properties of Trigonometric Inverse and Algebraic Inverse.  As clear by name Inverse means the opposite in position, directions, etc.

In mathematical language, it is defined as a reciprocal quantity.

## Trigonometric Inverse Tips and Tricks and Shortcuts:

They are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions.

### Property of Trigonometry inverse functions

PROPERTY 1

a) sin-1($\frac{1}{x }$) = cosec -1 x,x≥1 or x≤ -1

b) cos-1($\frac{1}{x }$) = sec -1x, x≥1 or x≤-1

c) tan -1($\frac{1}{x }$) = cot -1x, x>0

PROPERTY 2

a) sin-1(-x) = -sin-1(x), x ∈ [-1,1]

b) tan-1 (-x) = tan-1 (x) , x ∈ R

c) cosec-1 (-x) = -cosec-1 (x), |x| ≥ 1

PROPERTY 3

a) cos-1 (-x) =π-cos-1 x, x ∈ [-1,1]

b)  sec-1 (-x) = π -sec-1 x, |x|≥ 1

c) cot-1 (-x) = π – cot-1 x, x ∈ R

PROPERTY 4

a) sin-1 x+ cos-1x $=\frac{π}{2}$ , x∈ [-1,1]

b) tan-1 x + cot-1 x = $\frac{π}{2}$, x∈ R

c) cosec-1 + sec-1x = $\frac{π}{2}$ , |x|≥ 1

PROPERTY 5

a) tan-1 x + tan-1 y = tan-1 ($\frac{(x+y)}{(1-xy)}$ ), xy< 1

b)tan-1 x- tan -1 y = tan -1 ($\frac{(x-y)}{(1+xy)}$ ),xy > -1

PROPERTY 6

a)  2 tan-1x = sin-1($\frac{(2x)}{(1+x^2)}$), |x|≤1

b)  2 tan-1x = cos-1($\frac{(1-x^{2} )}{(1+x^{2})}$) , x≥0

c) 2 tan-1x = tan -1($\frac{(2x )}{(1-x^ 2)}$ ), -1<x<1

## Algebraic Inverse Tips and Tricks and Shortcuts:

Inverse is a Reverse of any quantity.

Addition is the opposite of subtraction; division is the opposite of multiplication, and so on.

For Example-

If, f is the inverse of y,

Then, the inverse of f(x)= 2x+3 can be written as,

f-1 (y)= $\frac{ (y-3)}{2} \$

### Inverse Questions and Answers :

Question 1 : Given $f(x) = \frac{3x}{4 – x}$

A. $f^{-1}(x) = \frac{4x}{3 + x}$
B. $f^{-1}(x) = \frac{4x}{x – 3}$
C. $f^{-1}(x) = \frac{4x}{3 – x}$
D. $f^{-1}(x) = \frac{4x}{3 + x}$

Solution:

For the first step we simply replace the function with a y
$y = \frac{3x}{4-x}$
Next, replace all the x’s with y’s and all the original y’s with x’s.
$x = \frac{3y}{4 – y}$
Solve the equation from Step 2 for y.
$x = \frac{3y}{4 – y}$
$x \times (4 – y) = 3y$
$4x – xy = 3y$
$4x = 3y + xy$
$4x = y(3 + x)$
$y = \frac{4x}{3 + x}$
Replace y with $f^{-1}$
$f^{-1}(x) = \frac{4x}{3 + x}$

Question 2 : Given $f(x) = \frac{7x}{3 – x}$

A. $f^{-1}(x) = \frac{3x}{7 – x}$
B. $f^{-1}(x) = \frac{7x}{3 + x}$
C. $f^{-1}(x) = \frac{3x}{7 + x}$
D. None of these

Solution:

For the first step we simply replace the function with a y
$y = \frac{7x}{3-x}$
Next, replace all the x’s with y’s and all the original y’s with x’s.
$x = \frac{7y}{3 – y}$
Solve the equation from Step 2 for y.
$x = \frac{7y}{3 – y}$
$x \times (3 – y) = 7y$
$3x – xy = 7y$
$3x = 7y + xy$
$3x = y(7 + x)$
$y = \frac{3x}{7 + x}$
Replace y with $f^{-1}$
$f^{-1}(x) = \frac{3y}{7 + y}$

Question 3 :  Let f(x) = (1 + 9x) / (4 – x). Find $f^{-1}(x)$.

A. $f^{-1}(x) = \frac{4 – x}{1 + 9x}$
B. $f^{-1}(x) = \frac{4x – 1}{9 + x}$
C. $f^{-1}(x) = \frac{1 – 9x}{4 + x)}$
D. $f^{-1}(x) = \frac{9x – 1}{x – 4}$

Solution:

To find $f^{-1}(x)$, we switch x and y and solve for y:
$x = \frac{1 + 9y}{4 – y}$,
x(4 – y) = 1 + 9y,
4x – xy = 1 + 9y,
y(9 + x) = 4x – 1,
$y = \frac{4x – 1}{9 + x}$
Therefore, $f^{-1}(x) = \frac{4x – 1}{9 + x}$

Question 4 : Let f(x) = 2x – 5. Find $f^{-1}(x)$.

A. $f^{-1}(x) = \frac{x}{2} – 5$
B. $f^{-1}(x) = \frac{x+5}{2}$
C. $f^{-1}(x) = \frac{2}{x} – 5$
D. $f^{-1}(x) = \frac{x-5}{2}$

Solution:

To find $f^{-1}(x)$, we switch x and y and solve for y:
x = 2y – 5,
x + 5 = 2y,
y = (x + 5) / 2.
Therefore, $f^{-1}(x) = \frac{x+5}{2}$

Question 5 : Let f(x) = 3x + 2. Find inverse of this function

A. (x – 2) / 3
B. (x + 2) / 3
C. (x – 3) / 2
D. (x + 3) / 2

Solution:

To find inverse of f(x), we switch x and y and solve for y:
x = 3y + 2,
x – 2 = 3y,
y = (x – 2) / 3.
Therefore, Inverse of f(x) = (x – 2) / 3.

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