Tips and Tricks and Shortcuts on Inverse

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 Tips and Tricks and Shortcuts

Question 1 .Prove that the inverse of an invertible odd function is also an odd function

Explanation

We know inverse of f is f^-1
f(f^-1(x)) = x

Now change x to-x
f(f^-1 (x)) = -(-x)

Now change -x in terms of f (f^-1(-x))
-f (f^-1(-x))= f (f^-1(-x))

which gives
f^-1(x) = -f ^-1(-x)

And hence proved f^-1 is also odd.

Question 2 Find the principal value of sin^-1 (- \frac{1}{2} \ )

Options

A)  ( \frac{1}{2} )

B) (- \frac{1}{2} )

C) ( \frac{π}{2} )

D) (- \frac{π}{6} )

Correct Answer: D

Explanation

let  sin^-1(- \frac{1}{2} ) = y,
then, sin y = (- \frac{1}{2} ) = sin ( \frac{π}{6} ) = sin (- \frac{π}{6})

We know that the range of the principal value branch of sin^-1 is (- \frac{π}{2}  ), ( \frac{π}{2} )

and sin (- \frac{π}{6} \ ) = ( \frac{1}{2} )

Therefore the principal value of sin^-1 (- \frac{1}{2} ) is (- \frac{π}{6} )

Read also:  Formulas for Inverse

                    How to Solve Inverse Questions Quickly

                    Inverse Questions and Answers