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Tricks of Inverse Trigonometric Function for JEE Mains & Advance Level Problems

Trick 1 : Writing the nth Term as Difference of two terms

Please see Examples
Q 2675080866

The sum of the infinite terms of the series `tan^(-1) (1/3)+tan^(-1) (2/9) + tan^(-1) (4/33) + ...........` is equal to

(A)

`pi/6`

(B)

`pi/4`

(C)

`pi/3`

(D)

`pi/2`

Solution:

`because T_r = tan^(-1) ( 2^(r-1)/(1+2^(2r-1)))`

` = tan^(-1) (( 2^r - 2^(r-1))/(1+2^r*2^(r-1)))`

` = tan^(-1) (2^r) - tan^(-1) (2^(r-1))`

`therefore S_n = underset(r = 1) overset(n)Sigma T_r = underset(r = 1) overset(n)Sigma tan^(-1) (2^r) - tan^(-1) (2^(r-1))`

` = tan^(-1) (2^n) - tan^(-1) (2^0)`

`= tan^(-1) (2^n) - pi/4`

Hence `S_(oo) = tan^(-1) (2^(oo)) - pi/4`

`= tan^(-1)(oo) - pi/4`

` = pi/2 - pi/4 = pi/4`
Correct Answer is `=>` (B) `pi/4`
Q 2625191061

The sum of the infinite series `cot^(-1) 2 + cot^(-1) 8 + cot^(-1) 18 + cot^(-1) 32 + ...` is equal to

(A)

`pi`

(B)

`pi/2`

(C)

`pi/4`

(D)

none of these

Solution:

`:. cot^(-1) 2 + cot^(-1) 8 + cot^(-1) 18 + cot^(-1) 32 + ...`

` T_n = cot^(-1) 2 n^2`

` = tan^(-1) (1/(2n^2))`

` = tan^(-1) ( 2/( 1+ 4n^2 - 1))`

` = tan^(-1) ( ( 2n + 1) - (2 n - 1))/( 1 + ( 2n + 1) ( 2n - 1))`

` = tan^(-1) ( 2n + 1) - tan^(-1) ( 2n - 1)`

`:. S = lim_(n->oo) sum_( n = 1)^n T_n`

` = lim_(n->oo) sum_( n = 1)^n {tan^(-1) (2 n + 1) - tan^(-1) (2n - 1)}`

` = lim_(n->oo) ( tan^(-1) ( 2n + 1) - tan^(-1) (2n -1)}`

` = tan^(-1) oo - tan^(-1) 1`

` = pi/2 - pi/4`

` = pi/4`
Correct Answer is `=>` (C) `pi/4`
Q 2665391265

The value of `tan^(-1) ( ( c_1 x - y)/( c_1 x + y) ) + tan^(-1) ((c_2 - c_1)/( 1 + c_2 c_1)) + tan^(-1) ((c_3 - c_2)/( 1 + c_3 c_2)) + ... + tan^(-1) (1/c_n)` is equal to

(A)

` tan^(-1) ( y/x)`

(B)

` tan^(-1) ( x/y)`

(C)

` - tan^(-1) ( x/y)`

(D)

None of these

Solution:

We have,

`tan^(-1) ( ( c_1 x - y)/( c_1 x + y) ) + tan^(-1) ((c_2 - c_1)/( 1 + c_2 c_1)) + tan^(-1) ((c_3 - c_2)/( 1 + c_3 c_2)) + ... + tan^(-1) (1/c_n)`

` = tan^(-1) ((x/y - 1/c_1)/( 1 + x/y 1/c_1)) + tan^(-1) ((1/c_1 - 1/c_2)/(1 + 1/(c_1c_2)))`

` tan^(-1) ((1/c_2 - 1/c_3)/(1 + 1/(c_2 c_3))) + .... + tan^(-1) (1/c_n)`

` = tan^(-1) (x/y) - tan^(-1) ( 1/c_1) + tan^(-1)( 1/c_1)`

` - tan^(-1) ( 1/c_2) + tan^(-1) ( 1/c_2) - tan^(-1) ( 1/c_3)`

` + .... - tan^(-1) ( 1/c_n) + tan^(-1) ( 1/c_n) = tan^(-1) ( x/y)`
Correct Answer is `=>` (B) ` tan^(-1) ( x/y)`
Q 2635491362

The sum of the infinite series `sin^(-1) (1/sqrt2) + sin^(-1) ( (sqrt2-1)/sqrt6)+sin^(-1) ( ( sqrt3 - sqrt2)/sqrt(12)) +......+.........+ sin^(-1) ( (sqrtn - sqrt(n-1))/sqrt({n(n+1)}))+ ...........` is

(A)

`pi/8`

(B)

`pi/4`

(C)

`pi/2`

(D)

`pi`

Solution:

`T_r = sin^(-1) ( ( sqrtr - sqrt((r-1)))/sqrt(r (r+1)))`

` = tan^(-1) ( ( sqrtr - sqrt(r-1))/(1+sqrtr sqrt(r-1)))`


`therefore S_n = underset(r = 1) overset(n) Sigma tan^(-1) ( ( sqrtr - sqrt(r-1))/(1+sqrtr sqrt(r-1)))`


` = S_n = underset(r = 1) overset(n) Sigma tan^(-1) sqrtr - tan^(-1) sqrt(r-1)`

` = tan^(-1) sqrtn - tan^(-1) sqrt0`

`= tan^(-1) sqrtn - 0`

`therefore S_(oo) =tan^(-1) oo = pi/2`
Correct Answer is `=>` (C) `pi/2`
Q 2665491365

If `a_1, a_2, a_3, ... , a_n` is an AP with common difference `d`, then ` tan [ tan^(-1) ( d/(1 + a_1a_2) ) + tan^(-1) ( d/(1 + a_2 a_3) ) + .... + tan^(-1) ( d/(1 + a_(n-1) a_n) ) ]` is equal to



(A)

` ((n - 1)d)/(a_1 + a_n)`

(B)

` ((n - 1)d)/(1 + a_1 a_n)`

(C)

` (n d)/( 1+ a_1 a_n)`

(D)

`(a_n - a_1)/(a_n + a_1)`

Solution:

` ∵ a_2 - a_1 = a_3 - a_2 = ..... = a_n - a_(n-1) = d`

` :. tan [ tan^(-1) ( d/(1 + a_1a_2) ) + tan^(-1) ( d/(1 + a_2 a_3) ) + .... + tan^(-1) ( d/(1 + a_(n-1) a_n) ) ]`

` = tan [ sum_(r=2)^n tan^(-1) ( d/( 1+ a_(r-1) a_r)) ]`

` = tan [ sum_(r=2)^n tan^(-1) ( (a_r - a_(r-1)) /( 1+ a_(r-1) a_r)) ]`

` = tan [ sum_(r=2)^n ( tan^(-1) a_r - tan^(-1) a_(r-1) ]`

`= tan (tan^(-1) a_n - tan^(-1) a_1 )`

` = tan ( tan^(-1) ((a_n - a_1)/(1 + a_1a_n)))`

` = (a_n - a_1)/(1 + a_1a_n) = ( a_1 + (n-1) d - a_1)/( 1 + a_1a_n)`

` = ((n-1)d)/(1 + a_1a_n)`
Correct Answer is `=>` (B) ` ((n - 1)d)/(1 + a_1 a_n)`

Trick 2 : Finding value just by looking identity

Please see Examples
Q 2605180968

If `sin^(-1) x +sin^(-1)y +sin^(-1)z = (3 pi)/2` and `f(1) = 2 , f(p+q) = f(p) * f(q) ∀ p , q in R` then `x^(f(1))+y^(f(2))+z^(f(3)) - ((x+y+z))/(x^(f(1))+y^(f(2))+z^(f(3)))` is equal to

(A)

`0`

(B)

`1`

(C)

`2`

(D)

`3`

Solution:

`because - pi/2 le sin^(-1) x le pi/2 , -pi/2 le sin^(-1) y le pi/2` and `-pi/2 le sin^(-1) z le pi/2`

Given `sin^(-1) x+sin^(-1)y +sin^(-1)z = (3pi)/2`

Which is possible only when

`sin^(-1) x = sin^(-1)y = sin^(-1)z = pi/2`

or ` x= y = z = 1`

put `p = q= 1`

then `f(2) = f(1) f(1) = 2*2 = 4` and put `p = 1 , q = 2`

then `f(3) = f(1) f(2) = 2*2^2 = 8`

`therefore x^(f(1))+y^(f(2))+z^(f(3)) - (x+y+z)/(x^(f(1))+y^(f(2))+z^(f(3)))`

` = 1+1+1 - 3/(1+1+1) = 3-1 = 2`
Correct Answer is `=>` (C) `2`
Q 2615191060

If ` sum_( i = 1)^(2n) sin^(-1) x_i = n pi ` , then ` sum_(i = 1)^(2n) `is equal to

(A)

`n`

(B)

`2n`

(C)

`(n ( n + 1))/2`

(D)

none of these

Solution:

`∵ sum_(i = 1)^(2n) sin^(-1) x_i = n pi = sum_(i = 1)^(2n) ( pi/2)`

Which is possible only when

` sin^(-1) x_i = pi/2 AA i`

` :. x_i = 1 AA i`

then ` sum_(i = 1)^(2n) x_i = sum_(i = 1)^(2n) 1 = 2n`
Correct Answer is `=>` (B) `2n`
Q 2645591463

If `cos^(-1) x + cos^(-1) y + cos^(-1) z = 3 pi`, then `x y + y z + z x` is equal to


(A)

`-3`

(B)

`0`

(C)

`3`

(D)

`-1`

Solution:

`∵ 0 <= cos^(-1) x <= pi , 0 <= cos^(-1) y <= pi`,

` 0 <= cos^(-1) z <= pi`

But given, `cos^(-1) x + cos^(-1) y + cos^(-1) z = 3pi`

Which is possible only when

`cos^(-1) x = pi , cos^(-1) y = pi` and `cos^(-1) z = pi`

`:. x = - 1, y = - 1` and `z = - 1`

Then, `xy + yz + zx = 3`
Correct Answer is `=>` (C) `3`

Trick 3 : Finding value of equation involving `cosx, tanx` by changing them in `tanx`

Please see examples
Q 2645391263

If `- 1 < x < 0`, then `sin^(-1) x` equals



(A)

`pi - cos^(-1) { sqrt( 1- x^2) }`

(B)

` tan^(-1) { x/sqrt(1-x^2)}`

(C)

` - cot^(-1) { sqrt(1 - x^2)/x }`

(D)

`cosec^(-1) x`

Solution:

`∵ -1 < x < 0,` then `- pi/2 < sin^(-1) x < 0`

Let ` sin^(-1) x = alpha , sin alpha = x`

Then, ` tan alpha = x/sqrt(1 - x^2)`

`:. alpha = tan^(-1) ( x/sqrt(1 - x^2) )`

`:. sin^(-1) x = tan^(-1) ( x/sqrt(1 - x^2) )`
Correct Answer is `=>` (B) ` tan^(-1) { x/sqrt(1-x^2)}`
Q 2625491361

The number of the positive integral solutions of ` tan^(-1) x + cos^(-1) (y/sqrt(1 + y^2)) = sin^(-1) ( 3/sqrt(10))` is



(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

Solution:

Since, ` tan^(-1) x + cos^(-1) (y/sqrt(1 + y^2)) = sin^(-1) ( 3/sqrt(10))`

` tan^(-1) x + tan^(-1) (1/y) = tan^(-1) 3`

or ` tan^(-1) (1/y) = tan^(-1) 3 - tan^(-1) x`

` = tan^(-1) ((3-x)/( 1 + 3x))`

or ` y = (1 + 3x)/(3 -x) = -3 + (10)/(3 -x)`

For positive integer `y, x` must be `1` and `2`

`:. y` (at `x = 1`) `= 2`

and `y` (at `x = 2`) `= 7`

Hence, solutions are `(x, y) = (1, 2), (2, 7)`
Correct Answer is `=>` (B) `2`
Q 2615691569

If the numerical value of `tan { cos^(-1) (4/5) + tan^(-1) (2/3) } ` is `a/b` then

(A)

`a+ b = 23`

(B)

`a- b = 11`

(C)

`3b =a+ 1`

(D)

`2a = 3b`

Solution:

Let `cos^(-1) (4/5) = alpha,` that is `cos alpha =4/5` , so that

`tan alpha = sqrt ({(5/4 )^2 -1}) =3/4`

(·:` 0 < alpha < pi` and `cos alpha > 0`)

and `tan(cos^(-1) (4/5) +tan^(-1) (2/3))`

`= (tan alpha + 2/3)/(1- tan alpha * 2/3)`

`=(3/4 + 2/3)/(1- 2/3 * 3/4)= 17/6 = a/b` (given)

So, `a=17,b=6,a+b=23,`

`a - b = 11` and `3b = a + 1`
Q 2650580414

The value of `tan [ cos^(-1) (4/5 )+ tan^(-1) (2/3)]` is

(A)

`6/17`

(B)

`7/16`

(C)

`16/7`

(D)

None of these

Solution:

Since `cos^(-1) (4/5) =tan^(-1) (3/4)`

`:. tan[cos^(-1) (4/5) + tan^(-1) (2/3)] =tan[ tan^(-1) 3/4 + tan^(-1) 2/3]= (3/4 + 2/3)/(1-3/4 * 2/3) =17/6`
Correct Answer is `=>` (D) None of these

Value of `(sin^(-1)x)^3 + (cos^(-1)x)^3`

See examples
Q 2615391269

The greatest and least values of `(sin^(-1) x)^3+ (cos^(-1) x)^3` are

(This question may have multiple correct answers)

(A) `(pi^3)/32`
(B) `- (pi^3)/8`
(C) `(7 pi^3)/8`
(D) `pi/2`
Solution:

Let `y= (sin^(-1) x)^3 +(cos^(-1) x)^3`

`= (sin^(-1) x + cos^(-1) x) { (sin ^(-1) x)^2 +(cos^(-1) x)^2- sin ^(-1)x cos^(-1) x}`

`=pi/2 { (sin ^(-1) x)^2 + (pi/2 - sin^(-1) x)^2 - sin^(-1) x(pi/2 - sin^(-1) x)}`

`=pi/2 {3 ( sin^(-1) x)^2 -(3 pi)/2 sin^(-1) x + (pi)^2/4}`

Since, `-1 le sin^(-1) x le 1`

or `-pi/2 le x le pi/2`

`y_(-pi/2) =(7 pi^2)/8`...............(i)

`y_(pi/2)=(pi^3)/8`................(ii)

and `3 (sin ^(-1) x)^2 -(3 pi)/2 sin^(-1) x+ (pi^2)/4 =(2 y)/pi`

`=> 3 (sin^(-1) x)^2 -(3 pi)/2 sin^(-1) x + (pi^2)/4 -(2 y)/pi =0`

For real, `B^2- 4AC ge 0`


`=> (9 pi^2)/4 - 3 pi^2 +(24 y)/pi ge 0`

`=> (-3 pi^2)/4 +(24 y)/ pi ge 0`

`:. y ge (pi^3)/32`..............(iii)

From Eqs. (i), (ii) and (iii),

Least value of `y` is `(pi^3)/32`

and greatest value of `y` is `(7 pi^3)/8`
Correct Answer is `=>` (A)
Q 2615491360

If `(tan^(-1) x)^2 + (cot^(-1) x)^2 = (5pi^2)/8` , then `x` equals



(A)

`0`

(B)

`-1`

(C)

`-2`

(D)

`-3`

Solution:

`(tan^(-1) x)^2 + ( pi/2 - tan^(-1) x)^2 = (5 pi^2)/8`

` => 2( tan^(-1) x)^2 - pi tan^(-1) x - ( 3 pi^2)/8 = 0`

` :. tan^(-1) x = ( pi pm sqrt( 4 pi^2))/4`

` = (3pi)/4 ` or ` - pi/4`

` :. x = -1`
Correct Answer is `=>` (B) `-1`

Some Beautiful Problems :

See examples
Q 2655491364

The solution of the equation `sin [2 cos^(-1){cot (2 tan ^(-1) x)}] = 0` are

(This question may have multiple correct answers)

(A) `pm 1`
(B) `1 pm sqrt 2`
(C) `-1 pm sqrt 2`
(D) None of these
Solution:

`sin [2 cos^(-1){cot (2 tan^(-1 x))}] = 0`

`=> 2 cos^(-1) {cot (2 tan^(-1) x)} = n pi, n in I`

`=> cos^(-1) (cot {2 tan^(-1) x)} =(n pi)/2 =0, pi/2 , pi`

`=> cot (2 tan^(-1) x) =1, 0,-1`

`=> cot {tan^(-1) ((2x)/(1-x^2)}= 1, 0,-1`

`=> cot {cot^(-1) ((1-x^2)/(2x)} =1,0,-1`

`=> (1-x^2)/(2x)=1,0,-1`

`=> 1-x^2= 2x, 0, -2x`

or `x^2 + 2x - 1 = 0, x^2- 1 = 0, x^2 - 2x - 1 = 0`

or, `x=-1 pm sqrt 2, x= pm 1, x= 1 pm sqrt 2`
Correct Answer is `=>` (A)
Q 2675491366

If `[ sin^(-1) cos^(-1) sin^(-1) tan^(-1) x] = 1, ` where `[.]` denotes the greatest integer function, then `x` belongs to the interval



(A)

`[tan sin cos 1, tan sin cos sin 1]`

(B)

`(tan sin cos 1, tan sin cos sin 1)`

(C)

`[- 1, 1]`

(D)

`[sin cos tan 1, sin cos sin tan 1]`

Solution:

We have, `1 le sin^(-1) cos^(-1) sin^(-1) tan^(-1) x le pi/2`

`=> sin1 le cos^(-1) sin^(-1) tan^(-1) x le 1`

`=> cos sin1 ge sin^(-1) tan^(-1) x ge cos1`

`=> sin cos sin1 ge tan^(-1) x ge sin cos1`

`=> tan cos sin1 ge x ge tan sin cos1`

`therefore x in [ tanx sin cos1 , tan sin cos sin1]`
Correct Answer is `=>` (A) `[tan sin cos 1, tan sin cos sin 1]`
Q 2620580411

The solution of the inequality `log _(1//x) sin^(-1) x > log_(1//x) cos^(-1) x ` is

(A)

`x in [ 0, 1/sqrt 2]`

(B)

` x in [ 1/sqrt 2 , 1]`

(C)

`x in (0, 1/sqrt 2)`

(D)

None of these

Solution:

`log _(1//x) sin^(-1) x > log _(1//x) cos^(-1) x`

`=> cos^(-1) x > sin^(-1) x , 0 < x < 1`

`=> cos ^(-1) x > pi/4 , 0 < x < 1`

`=> 0 < x < 1/sqrt 2`
Correct Answer is `=>` (C) `x in (0, 1/sqrt 2)`
Q 2650680514

The sum of solutions of the equation `2 sin^(-1) sqrt (x^2 +x+1) + cos^(-1) sqrt(x^2+x)= (3 pi)/2` is

(A)

`0`

(B)

`-1`

(C)

`1`

(D)

`2`

Solution:

`0 le x^2 + x+1 le 1` and `0 le x^2 + x le 1`

`:. x=-1,0`

for `x=-1`

L.H.S `= 2 sin^(-1) 1 + cos^(-1) 0 = (3 pi)/2`

`:. x=-1` is a solution.

for `x=0` L.H.S `=2 sin^(-1) 1 + cos^(-1) 0 =(3 pi)/2`

`:. x=0` is a solution

`:.` sum of the solution `=-1`
Correct Answer is `=>` (B) `-1`
Q 2662191935








Column IColumn II
(A) The number of possible values of `k` if fundamental
period of `sin^(-1) (sin kx)` is `pi/2` , is 		(P) `1`
(B) Numbers of elements in the domain of
`f(x) = tan^(-1) x + sin^(-1) x + sec^(-1 ) x` is 		(Q) `2`
(C) Period of the function `f(x) = sin ((pi x)/2) * cosec ((pi x)/2)` is (R) `3`
(D) If the range of the function `f(x) = cos^(-1) [5x]` is `{a, b, c}`
and `a + b + c = (lambda pi )/2`, then `lambda` is equal to
(where `[*]` denotes greatest integer) 		(S) `4`
(T) `0`


(A)

(A)-> (p), (B)-> (q), (C)-> (q), (D)-> (r)

(B)

(A)-> (s), (B)-> (q), (C)-> (q), (D)-> (r)

(C)

(A)-> (q), (B)-> (q), (C)-> (q), (D)-> (r)

(D)

(A)-> (t), (B)-> (q), (C)-> (q), (D)-> (r)

Solution:

(A) Fundamental period of `sin^(-1) (sin kx)` is `(2 pi )/(|k |) = pi/2` , i.e., `| k | =4` i.e., `k= pm 4`

(B) domain of `tan^(-1) x` is `R`, domain of `sin^(-1) x is [-1, 1]`, domain of `sec^(-1) x` is `(-oo, -1] cup [ 1, oo)`

`:. ` domain of `f(x)` is `{-1, 1}`

(C) `pi` is a period of `sin x * cosec x`

`:. pi xx 2/pi ` is a period of `sin (pi x)/2 * cosec (pi x)/2`

i.e., `2` is a period of `sin (pi x)/2 cosec (pi x)/2`

`f(x+1) = sin pi/2 (x+1) cosec pi/2 (x+1) = cos pi/2 x * sec pi/2 x ne f(x)`

(D) `f(x) = cos^(-1) [5x]`

`[5x]` can take the values `- 1, 0, 1`

`:.` range `= {pi , pi/2 , 0 } `

`:. a+b +c = pi + pi/2 + 0 = (3 pi)/2`
Correct Answer is `=>` (C) (A)-> (q), (B)-> (q), (C)-> (q), (D)-> (r)
 
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