2.6 Supplementary reading for JEE Main and Advance

Please Note that this part of the text has been taken from the previous edition of NCERT Class 12 Mathematics Textbook.

Accessible NCERT Class 12 Mathematics Textbook for Blind and Visually Impaired Students facilitated by Professor T K Bansal.

Same is true for other five inverse trigonometric functions as well.

We will now prove some properties of inverse trigonometric functions.

Property 1

\[i.\ \sin^{−1}(\frac{1}{x})\ =\ \csc^{−1} x,\ x\ ≥\ 1\ ∨\ x\ ≤\ −1\]


\[ii.\ \cos^{−1}(\frac{1}{x})\ =\ \sec^{−1} x,\ x\ ≥\ 1\ ∨\ x\ ≤\ −1\]


\[iii.\ \tan^{−1}(\frac{1}{x})\ =\ \cot^{−1} x,\ x\ >\ 0\]

To prove the first result,


we put cosec^−1 x = y,


i.e., x = cosec y


Therefore 1/x = sin y


Hence sin^−1 (1/x) = y


or sin^−1 (1/x) = cosec^−1 x

Similarly, we can prove the other parts.

Property 2

\[i.\ \sin^{−1} (−x)\ =\ −\ \sin^{−1} x,\ x\ ∈\ [−1,\ 1]\]


\[ii.\ \tan^{−1}(−x)\ =\ −\ \tan^{−1} x,\ x\ ∈\ R\]


\[iii.\ \csc^{−1}(−x)\ =\ −\ \csc^{−1}x, |x|\ ≥\ 1\]

To prove these:

Let sin^−1 ( −x) = y,


i.e., −x = sin y


so that x = − sin y,


i.e., x = sin ( −y).


taking inverse on both sides, we get:


sin^−1 x = − y


= − sin^−1 ( −x)

Therefore sin^−1 (−x)


= − sin^−1 x

Similarly, we can prove the other parts.

Property 3

\[i.\ \cos^{−1}(−x)\ =\ π\ −\ \cos^{−1}x,\ x\ ∈\ [−1,\ 1]\]


\[ii.\ \sec^{−1}(−x)\ =\ π\ −\ \sec^{−1}x,\ |x|\ ≥\ 1\]


\[iii.\ \cot^{−1}(−x)\ =\ π\ −\ \cot^{−1}x,\ x\ ∈\ R\]

To prove the above:

Let, cos^−1 (−x) = y


i.e., − x = cos y


so that x = − cos y = cos ( π − y)


taking inverse on both sides:


Therefore cos^−1 x = π − y = π − cos^−1 (−x)


Hence cos^−1 (−x) = π − cos^−1 (x)

Similarly, we can prove the other parts.

Property 4

\[i.\ \sin^{−1}x\ +\ \cos^{−1}x\ =\ \frac{π}{2},\ x\ ∈\ [−1,\ 1]\]


\[ii.\ \tan^{−1}x\ +\ \cot^{−1}x\ =\ \frac{π}{2},\ x\ ∈\ R\]


\[iii.\ \csc^{−1}x\ +\ \sec^{−1}x\ =\ \frac{π}{2},\ |x|\ ≥\ 1\]

To prove these results:

Let sin^−1 (x) = y.


Then x = sin y = cos ((π/2) − y)


Therefore cos^−1 (x)= (π/2) − y


= π/2 − sin^−1 (x)


Hence sin^−1 (x) + cos^−1 (x) = π/2

Similarly, we can prove the other parts.

Property 5

\[i.\ \tan^{−1}x\ +\ \tan^{−1}y\ =\ \tan^{−1} \frac{x+y}{1−xy},\ xy\ <\ 1\]

\[ii.\ \tan^{−1} x\ −\ \tan^{−1} y\ =\ \tan^{−1} \frac{x−y}{1+xy},\ xy\ >\ −1\]

\[iii.\ \tan^{−1}x\ +\ \tan^{−1}y\ =\ π\ +\ \tan^{−1}\frac{x+y}{1−xy},\ xy\ >\ 1;\ x,\ y\ >\ 0\]

To prove these results:

Let tan^−1 (x) = θ and tan^−1( y) = φ.


Then x = tan θ, and y = tan φ


We know, tan(θ + φ)


= (tan θ + tan φ) / (1 − tan θ tan φ)


= (x + y)/(1 − xy)

This gives (θ + φ) = tan^−1 ((x + y)/(1 − xy))


Hence tan^−1 x + tan^−1 y = tan^−1 ((x + y)/(1 − xy))

In the above result, if we replace y by − y, we get the second result and by replacing y by x, we get the third result as given below.

Property 6

\[i.\ 2 \tan^{−1}x\ =\ \sin^{−1} \frac{2x}{1+x^2},\ |x|\ ≤\ 1\]


\[ii.\ 2 \tan^{−1}x\ =\ \cos^{−1} \frac{1−x^2}{1+x^2},\ x\ ≥\ 0\]


\[iii.\ 2 \tan^{−1} x\ =\ \tan^{−1} \frac{2x}{1−x^2}, −1\ <\ x\ <\ 1\]

To prove these results:

Let tan^−1 (x) = y,


then x = tan y.

Now


sin^−1 (2x/(1 + x^2))


= sin^−1 (2 tan y /(1 + tan^2 y))


= sin^−1 (sin 2y)


= 2y


= 2tan^−1 (x)

Also,


cos^−1 (1 − x^2)/(1 + x^2)


= cos^−1 (1 − tan^2(y))/(1 + tan^2(y))


= cos^−1 (cos 2y) = 2y


= 2tan^−1 x

(iii) Can be worked out similarly.


We now consider some examples.

Example 7

Show that


\[\tan^{−1}\frac{1}{2}\ +\ \tan^{−1}\frac{2}{11}\ =\ \tan^{−1} \frac{3}{4}\]

Solution:

By property 5 (i), we have


L.H.S. = tan^−1 (1/2) + tan^−1 (2/11)


= tan^−1 [(1/2 + 2/11)]/[(1 − 1/2) × (2/11)]


= tan^−1 (15/20) = tan^−1 (3/4) = R.H.S.

Example 8

Prove that


\[\tan^{−1}x\ +\ \tan^{−1}\frac{2x}{1}\ −\ x^2 \]


= tan^−1 ((3x − x^3)/(1 − 3x^2)), |x| < 1/√3

Solution :

Let x = tan θ.


Then θ = tan^−1 (x).


We have


R.H.S. = tan^−1 ((3x − x^3)/(1 − 3x^2))


= tan^−1 ((3tan θ − tan^3θ)/(1 − 3 tan^2θ))


= tan^−1 (tan3 θ) = 3 θ


= 3tan^−1 (x)


= tan^−1 (x) + 2 tan^−1 (x)


= tan^−1 (x) + tan^−1 ((2x)/(1 − x^2))


= L.H.S. (Why?)

Example 9

Find the value of


\[\cos {\sec^{−1}x\ +\ \csc^{−1} x},\ |x|≥1\]

Solution:

We have cos (sec^−1 (x) + cosec^−1 (x))


= cos (π/2) = 0.

Example 10

Show that


\[\sin^{−1} \frac{3}{5}\ −\ \sin^{−1} \frac{8}{17}\ =\ \cos^{−1} \frac{84}{85}\]

Solution :

Let sin^−1 (3/5) = x and sin^−1 (8/17) = y


Therefore sin x = (3/5) and sin y = (8/17)


Now cos x = √(1 − sin^2 x) = √(1 – (9/25)) = 4/5 (Why?)


and cos y = √(1 − sin^2 (y))


= √(1 – (64/289)) = 15/17


We have cos (x−y) = cos x cos y + sin x sin y


= (4/5) × (15/17) + (3/5) × (8/17) = (84/85)


Therefore x − y = cos^−1 (84/85)


Hence sin^−1 (3/5) − sin^−1(8/17)


= cos^−1 (4/85)


Show that sin^−1 (3/5) − sin^−1 (8/17) = cos^−1 (84/85)

Example 11

Show that


\[\sin^{−1} \frac{12}{13}\ +\ \cos^{−1} \frac{4}{5}\ +\ \tan^{−1} \frac{63}{16} =\ π\]

Solution:

Let sin^−1 (12/13) = x, cos^−1(4/5) = y, tan^−1 (63/16) = z


Then sin x = (12/13), cos y = (4/5), tan z = (63/16 )


Therefore cos x = (5/13), sin y = (3/5), tan x = (12/5) and tan y = (3/4)


We have tan (x + y) = (tan x + tan y)/(1 – tan x tan y)


= [(12/5) + (3/4)]/[(1 – (12/5) × (3/4)] = − (63/16)

Hence tan(x + y) = − tan z


i.e., tan (x + y) = tan (−z) or tan (x + y) = tan (π − z)


Therefore (x + y) = − z or (x + y) = (π – z)


Since x, y and z are positive, (x + y) ≠ − z (Why?)


Hence (x + y + z) = π or sin^−1 (12/13) + cos^−1 (4/5) + tan^−1 (63/16) = π

Example 12

Simplify


\[\tan^{−1} \frac{a \cos x\ −\ b \sin x}{b \cos x\ +\ a \sin x},\ \frac{a}{b} \tan x\ >\ −1\]

Solution:

We have, tan^−1 [(a cos x − b sinx)/(b cosx + a sinx)]


= tan^−1 [{(a cosx − b sinx)/ (b cosx)}/{(b cosx + a sinx)/(b cosx)}]


= tan^−1 [(a/b – tan x)/(1 + (a/b) tanx)]


= tan^−1 (a/b) − tan^−1 (tan x) = tan^−1 (a/b) − x

Example 13

Solve


\[\tan^{−1} 2x\ +\ \tan^{−1} 3x\ =\ \frac{π}{4}\]

Solution:

We have tan^−1 (2x) + tan^−1 (3x) = (π/4)


or tan^−1 ((2x + 3x)/(1 − 2x × 3x)) = π/4


i.e., tan^−1 ((5x)/ (1 − 6x^2)) = (π/4)


Therefore (5x)/(1 − 6x^2) = tan( π/4) = 1


or 6x^2 + 5x − 1 = 0


i.e., (6x − 1) (x + 1) = 0


which gives x = (1/6) or x = − 1.

Since x = − 1 does not satisfy the equation, as the L.H.S. of the equation becomes negative, x = (1/6) is the only solution of the given equation.