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1.

Prove that:`sin^(-1)(12)/(13)+cos^(-1)4/5+tan^(-1)(63)/(16)=pi`

Answer» here, `alpha + beta + gamma = pi`
`sin alpha = 12/13`
so, `tan alpha = 12/5`
and `cos beta = 4/5`
so, `tan beta= 3/4`
and `tan gamma = 63/16`
so,`alpha + beta = pi -gamma`
`tan(alpha + beta) = tan(pi- gamma)`
`(tan alpha + tan beta)/(1 - tan beta) = (tan pi -tan gamma)/(1+ tan pi tan gamma) `
`= (12/5 + 3/4)/(1- 12/5 xx 3/4) = -(63/16)`
`(48 + 15)/(-16) = -63/16`
hence proved
Discussion
2.

What is greater, `tan1 or tan^(-1)1?`

Answer» `tan 1 or tan^-1 1`
`1> pi/4`
`1> tan^-1 1`
applying tan on both sides for above eqn
`tan 1 > tan pi/4`
`tan 1 > 1`
now, `tan 1 > tan^-1 1`
`tan 1 > 1 > tan^-1 1`
Answer
Discussion
3.

In a ` A B C ,`, if C is a right angle, then `tan^(-1)(a/(b+c))+tan^(-1)(b/(c+a))=``pi/3`(b) `pi/4`(c) `(5pi)/2`(d) `pi/6`

Answer» `tan^-1 x + tan^-1 y = tan^-1 ((x+y)/(1-xy))`
so, `tan^-1 (a/(b+c)) + tan^-1(b/(a+c))`
`tan^-1((a/(b+c) + b/(a+c))/(1- (ab)/((a+c)(b+c))))`
`= tan^-1[ (a^2 + ac + b^2 + bc)/(ab + bc + ac + c^2 - ab)]`
`= tan^-1[(a^2+ b^2 + ac+bc)/(bc + ac+c^2)]`
so, `/_ c= pi/2`
`cos c = (a^2 + b^2 - c^2)/(2ab)=0`
`a^2 + b^2 = c^2`
`tan^-1 [ (bc+ac+c^2)/(bc + ac+c^2)]`
`= tan^-1 1 = pi/4`
Answer
Discussion
4.

Prove that:`(i)tan^(-1){(sqrt(1+cosx)+sqrt(1-cosx))/(sqrt(1+cosx)-sqrt(1-cosx))}=pi/4+x/2`,

Answer» `cos 2 theta = 2 cos^2 theta - 1`
`1 + cos 2 theta = 2 cos^2 theta`
so, `1 + cos x = 2cos^2 (x/2)`
Also `cos 2 theta = 1- 2sin^2 theta`
`1- cos 2 theta = 2 sin^2 theta`
so, `1 - cosx = 2sin^2(x/2)`
now, putting it in the equation given
`tan^-1{(sqrt(2cos^2(x/2)) - sqrt(2sin^2(x/2)))/(sqrt(2cos^2(x/2)) - sqrt(2sin^2(x/2)))}`
`= tan^-1{ (sqrt2 cos(x/2) + sqrt2 sin(x/2))/(sqrt2 cos(x/2) - sqrt2sin(x/2))}`
`= tan^-1{(1+tan(x/2))/(1- tan(x/2))}`
`= tan^-1 {(tan (pi/2) + tan(x/2))/(1 - tan(pi/4) tan(x/2))}`
`= tan^-1{tan(pi/4 + x/2)}`
`= pi/4 + x/2 ` = RHS
hence proved
Discussion
5.

Prove that `tan(cot^(-1)x)=cot(tan^(-1)x)`

Answer» `tan(cor^-1 x) = cot(tan^-1 x) `
`tan^-1 x + cot^-1 x = pi/2`
`tan theta = cot(pi/2 - theta)`
LHS: `tan(cot^-1 x) = cot(pi/2 - cot^-1 x)`
`= cot(tan^-1 x) `=RHS
hence proved
Discussion
6.

Write the value of `sin^-1(sin1550^@)`

Answer» `1550^@ = 1550 xx pi/180`
`= (155 pi)/18`
`=> sin^-1(sin ((155 pi)/18))`
`= sin^-1 (sin (8 pi + (11 pi)/18))`
`=> sin^-1(sin((11pi)/18))`
`= sin^-1[sin(pi- (7pi)/18)]`
`=> sin^-1(sin((7pi)/18))= (7pi)/18`
Answer
Discussion
7.

Prove that:`tan^(-1)((1-x^2)/(2x))+cot^(-1)((1-x^2)/(2x))=pi/2`

Answer» `cot^-1 (1/x) = tan^-1 x`
`tan^-1((1-x^2)/(2x)) + tan^-1((2x)/(1-x^2))`
as we know `tan^-1 x + tan^-1 y = tan^-1((x+y)/(1-xy))`
so, `tan^-1(((1-x^2)/(2x) + (2x)/(1-x^2))/(1- (1-x^3)/(2x)xx(2x)/(1-x^2)))`
`= tan^-1(oo) = pi/2`
hence proved
Discussion
8.

Write the value of `cos^2(1/2cos^(-1)3/5)`

Answer» `cos 2 theta = 2cos^2 theta- 1`
`=> cos^2 theta = (1 + cos 2 theta)/2`
`cos^2 (1/2 cos^-1(3/5))`
`= (1 + cos(2 xx 1/2 cos^-1 (3/5)))/2`
`=> (1 + cos(cos^-1(3/5)))/2`
`= (1 + (3/5))/2 = 8/(2 xx 5)`
`= 4/5`
Answer
Discussion
9.

Solve the equation: `cos^(-1)(a/x)-cos^(-1)(b/x)=cos^(-1)(1/b)-cos^(-1)(1/a)`

Answer» `cos^-1 (a/x) + cos^-1 (1/a) = cos^-1(b/x) + cos^-1 (1/b) `
`cos^-1 x + cos^-1 y = cos^-1 (xy - sqrt(1-x^2) sqrt(1-y^2))`
`cos^-1 (a/x 1/a - sqrt(1- a^2/x^2)sqrt(1- 1/a^2))`
`cos^-1 (b/x*1/b - sqrt(1- b^2/x^2) sqrt(1- 1/b^2))`
`(1- a^2/x^2) (1-1/a^2) = (1- b^2/x^2)(1- 1/b^2) `
`= 1- 1/a^2 - a^2/x^2 + 1/x^2 = 1- 1/b^2 - b^2/x^2 + 1/x^2`
`a^2/x^2 - b^2/x^2 = 1/b^2 - 1/a^2`
`= 1/x^2 [ a^2- b^2] = (a^2- b^2)/(a^2b^2) `
`x^2 = (ab)^2`
`x= ab`
Answer
Discussion
10.

If `tan^(-1)x+tan^(-1)y=pi/4,`then write the value of `x+y+x ydot`

Answer» `tan^-1 x + tan^-1 y = pi/4`
`tan^-1((x+y)/(1-xy)) = pi/4`
`(x+y)/(1-xy) = tan (pi/4)`
`(x+y)/(1-xy) = 1`
`x+y = 1-xy`
`x+y+xy= 1`
Answer
Discussion
11.

Evaluate:`sin(cos^(-1)3/5+cos e c^(-1)(13)/5)`

Answer» `sin(cos^-1 (3/5) + cosec^-1(3/5))`
`= sin(cos^-1(3/5)) cos(cosec^-1(13/5)) + cos(cos^-1(3/5))sin(cosec^-1(13/5))`
`= sin (sin^-1(4/5)) cos(cos^-1(12/13)) + cos(cos^-1(3/5))sin(sin^-1(5/13))`
here, `cos theta = 3/5`
`sin theta = 4/5`
so, in equation
`=4/5 xx 12/13 + 3/5 + 5/13`
`= 4/65 + 15/65 = 63/65`
Answer
Discussion
12.

Prove that:`"sin"[cot^(-1){"cos"(tan^(-1)x)}]=sqrt((x^2+1)/(x^2+2))````cos"[tan^(-1){"sin"(cot^(-1)x)}]=sqrt((x^2+1)/(x^2+2))``

Answer» LHS= `sin{cot^-1 { cos(tan^-1x)}}`
let `tan^-1 x = theta`
`tan theta = x`
`cos theta = 1/(sqrt(1+x^2))`
now, `sin^-1[ cot^-1 (1/sqrt(1+x^2))]`
`= siny`
`= sqrt(x^2 + 1)/sqrt(x^2 + 2) = `RHS
2) `cos[tan^-1{sin(cot^-1 x)}] `
`cos[tan^-1{sin theta}] = cos[ tan^-1( 1/sqrt(x^2+1))]`
`= cos y`
`sqrt(x^2 + 1)/sqrt(x^2+2) =RHS`
hence proved
Discussion
13.

Evaluate the following:`sin^(-1)(sin 10)`

Answer» range will be `[-pi/2, pi/2]`
`[-1.57, 1.57]`
`1^@ -> 57^@`
`10^@ -> ?`
`10^@` belongs to 3rd quad
ans should be negative
`10^@ - pi = 6.78`
`10 - 2 pi = 3.72`
`10 - 3pi = 0.58`
`sin^-1(sin(3pi-(0))`
`= -0.58`
Answer
Discussion
14.

If `cos e c^(-1)x+cos e c^(-1)y+cos e c^(-1)z=-(3pi)/2,`find the value of `x/y+y/z+z/xdot`

Answer» `cosec^-1 x=> [ -pi/2,0) uu (0, pi/2]`
`cosec^-1 x = -pi/2`
`cosec^-1 y = -pi/2`
`cosec^-1 z = -pi/2`
`x= -1`
`y= -1`
`z=-1`
`=> x/y + y/z + z/x`
`-1/-1 + -1/-1 + -1/-1`
= `1 + 1 + 1 = 3`
Answer
Discussion
15.

`"If"sin{cot^(-1)(x+1)}=cos(tan^(-1)x),`then find `x.`

Answer» `sin{cost^-1 (x+1)} = cos(tan^-1 x)`
`cot^-1(x+1) = sin^-1( 1/sqrt(1+ (x+1)^2))`
`tan^-1 x = cos^-1 (1/sqrt(1+x^2))`
=`sin[sin^-1 1/(sqrt(1+ (x+1)^2))] = cos[ cos^-1 (1/sqrt(1+x^2))]`
`= 1/sqrt(x^2 + 2x+ 2) = 1/(1+x^2) `
`x^2 + 1 = x^2 + 2x+2`
`2x+ 1 = 0`
`x= -1/2`
Answer
Discussion
16.

Prove that : `tan^(-1)(2/11)+tan^(-1)(7/24)=tan^(-1)(1/2)`

Answer» here, `alpha + beta = gama`
`tan alpha = 2/11`
`tan beta = 7/24`
as,`tan(alpha+ beta) = tan gamma`
`=> (tan alpha + tan beta)/(1 - tan alpha tan beta) `
`=(2/11 + 7/24)/(1- 2/11 xx 7/24)`
`= (48 + 77)/(24 xx 11 -14) = 125/250 = 1/2`
now, `tan(alpha + beta) = 1/2= tan gamma `
`gamma = tan^-1 1/2`
hence proved
Discussion
17.

If `(sin^(-1)x)^2+(cos^(-1)x)^2=(17pi^2)/(36)`, find `xdot`

Answer» `(17 pi^2)/36 = (sin^-1 x + cos^-1 x)^2 - 2 sin^-1 x cos^-1 x`
`(17 pi^2)/36 = pi^2/4 - 2sin^-1 x[ pi/2 - sin^-1 x]`
`2(sin^-1 x)^2 - pi(sin^-1 x) + pi^2/4- 17 pi^2/36= 0`
`2(sin^-1 x)^2 - pi(sin^-1 x) - (8pi^2)/36 = 0`
`(sin^-1 x)^2 - pi/2 (sin^-1 x) - pi^2 / 9= 0`
`sin^-1 x = (pi/2 +- sqrt(pi^2/4 + (4 pi^2)/9))/(2(1))`
`sin^-1 x = (pi/2 +- sqrt((25 pi^2)/(36)))/2`
now, `(pi/2 + 5pi/6)/2 or (pi/2 - 5 pi/6)/2`
`8pi/24 = pi/3or -pi/3`
`sin^-1 x = pi/3`
`x = sqrt3/2`
and `sin^-1 x = -pi/6`
`x= -1/2`
Answer
Discussion
18.

Find the maximum and minimum values of `(sin^(-1)x)^3+(cos^(-1)x)^3,`where `-1lt=xlt=1.`

Answer» `(sin^-1 x)^3 + (cos^-1 x)^3 = (sin^-1 x + cos^-1 x)^3 - 3sin^-1 x cos^-1 x (sin^-1x + cos^-1 x)`
`y = (pi/2)^3 - 3sin^-1 x (pi/2 - sin^-1 x )*(pi/2)`
`y= pi^3/8 - 3pi^2/4 sin^-1 x + 3 pi/2 (sin^-1 x)^2`
`3 pi/2 theta^2 - 3 pi^2/4 theta + pi^3/8 -y=0 `
`(theta - pi/4)^2 - pi^2/16 + pi^2/12 -2y/(3y) = 0`
`(theta- pi/4)^2 + pi^2/48 - (2y)/(3 pi) = 0`
`-pi/2 <= sin^-1 x <= pi/2`
`-(3pi)/4 <= sin^-1 x - pi/4 <= pi/4`
`0 <= (sin^-1 x - pi/4)^2 <= (9pi^2)/(16)`
`pi^2/48 <= 24/(3 pi) <= (9pi^2)/16 + pi^2/48`
`y max= (7pi^2)/8 ; y min = pi^3/32`
Answer
Discussion
19.

Prove that:`tan^(-1)((1-x)/(1+x))-tan^(-1)((1-y)/(1+y))=sin^(-1)((y-x)/(sqrt((1+x^2) (1+y^2))))`

Answer» `tan^-1( (1-x)/(1+x)) - tan^-1 ((1-y)/(1+y))`
`tan^-1 (1) - tan^-1 x - [tan^-1 (1) - tan^-1 y]`
`= tan^-1 y - tan^-1 x`
`= tan^-1 ((y-x)/(1+xy))`
`= sqrt(y^2 +x^2 -2xy+1+ x^2y^2 + 2xy)`
`= sqrt(y^2(1+x^2)+1(1+x^2))`
`= sqrt((y^2+1)(1+x^2)`
`sin^-1 ((y-x)/(sqrt(1+x^2)sqrt(1+y^2)))`
answer
Discussion
20.

Simplify each of the following:`cos^(-1)(3/5cosx+4/5sinx)`, where `-(3pi)/4lt=xlt=pi/4`

Answer» `cos^-1 (3/5 cos x + 4/5 sin x)`
here, `cos alpha= 3/5`
`sin alpha = 4/5`
so,`cos^-1[cos alpha cos x + sin alpha sin x]`
`= cos^-1[ cos(x- alpha)]`
`x- alpha`
as`alpha = cos^-1 (3/5) = sin^-1(4/5) = tan^-1(4/3)`
so,answer=`x- tan^-1 (4/3)`
Discussion
21.

Evaluate : `cos(2cos^(-1)x+sin^(-1)x)a tx=1/5`

Answer» `cos(2cos^-1 x + sin^-1 x) `at `x=1/5`
`=> sin^-1 x + cos^-1 x = pi/2`
`(cos^-1x)= sqrt(1-x^2)`
`=> cos(2cos^-1 x + sin^-1x) = cos(cos^-1 x+ (cos^-1 x + sin^-1 x))`
`=> cos(pi/2 + cos^-1 x) = - sin(cos^-1 x )`
`= sin (cos^-1(1/5))`
`= - sqrt(1- (1/5)^2)`
`= sqrt(24/25) = -2sqrt6/5`
Answer
Discussion
22.

Write each of the following in the simplest form:`cot^(-1){a/(sqrt(x^2-a^2))},|x|> a`

Answer» `cot^-1{a/sqrt(x^2 - a^2)}`
let `x= acosec theta`
`cot^-1 { a/sqrt(a^2 cosec^2 theta - a^2)}`
`cot^-1 { a/(a sqrt(cosec^2 theta -1))}`
`cot^-1 (tan theta)`
`cot^-1(cot(pi/2 - theta))`
`pi/2 - cosec^-1 (x/a) `
`= sec^-1 (x/a)`
Answer
Discussion
23.

Write the principal value of `cos^(-1)(cos680^0)dot`

Answer» `cos^-1(cos 680^@) `
As, `680= 360 + 320`
so, `cos^-1[cos(360+320)]`
`cos^-1 [ cos(360 - 40)] `
`cos^-1[cos(40)]`
`cos^-1[cos(40 xx pi/100)]`
`cos^-1[cos((2pi)/9)]`
`(2pi)/9`
Answer
Discussion
24.

Evaluate each of the following:`cot^(-1)1/(sqrt(3))-cos e c^(-1)(-2)+sec^(-1)(2/(sqrt(3)))``cot^(-1){2"cos"(sin^(-1)(sqrt(3))/2)}``cos e c^(-1)(-2/(sqrt(3)))+2cot^(-1)(-1)``tan^(-1)(1/(sqrt(3)))+cot^(-1)(1/(sqrt(3)))+tan^(-1)(sin(-pi/2))`

Answer» (i) `cot^-1 (1/sqrt3) - cosec^-1(-2) + sec^-1(2/sqrt3)`
`= pi/3 - (-pi/6)+ pi/6= pi/3 + pi/6 + pi/6 = (2pi)/3`
(ii) `cot^-1(2cos(sin^-1 (sqrt3/2)))`
`= cot^-1 { 2cos(pi/3)}`
`= cot^-1{1}= pi/4`
(iii) `cosec^-1 (-2/sqrt3) + 2cot^-1(-1)`
`= -pi/6 + 2(-pi/4) = - pi/6 + (-pi/2)`
`= -(2pi)/3`
(iv) `tan^-1(1/sqrt3) + cot^-1(1/sqrt3) + tan^-1(sin(-pi/2))`
as `tan^-1 x + cot^-1 x = pi/2`
so `pi/2 + tan^-1(-1)`
`pi/2 - pi/4 = pi/4`
Answers
Discussion
25.

Evaluate:`tan{1/2cos^(-1)(sqrt5/3)}`

Answer» `tan[ 1/2 cos^-1(sqrt5/3)]`
`1/2 cos^-1(sqrt5/3) = theta`
`cos^-1(sqrt5/3) = 2 theta`
`cos 2theta = sqrt5/3`
`(1- tan^2 theta)/(1 + tan^2 theta) = sqrt5/3`
`3- 3 tan^2 theta = sqrt5 + sqrt5 tan^2 theta`
`sqrt((3 - sqrt5)/(3 + sqrt5))= tan theta`
`sqrt(((3- sqrt5)(3- sqrt5))/((3 + sqrt5)(3- sqrt5))) = tan theta`
`tan theta = sqrt((3- sqrt5)^2/4)`
`= (3- sqrt5)/2`
Answer
Discussion
26.

If `y=cot^(-1)(sqrt(cos"x"))-tan^(-1)(sqrt(cos"x")),`prove that `siny=tan^2x/2`

Answer» `y= cot^-1 sqrt(cos x) - tan^-1 sqrt cosx`
`y = cot^-1 sqrtcosx - tan^-1 sqrt cos x `
`= tan^-1 sqrt secx - tan^-1 sqrt cos x`
`y = tan^-1 ((sqrt secx - sqrt cos x)/(1+1)) `
`= tan^-1 ( (sqrt secx - sqrt cosx)/2)`
`y= tan^-1 ((1 -cosx)/(2 sqrtcosx))`
`tan y = (1- cosx)/(2 sqrt cosx)`
now, `sin y = (1 -cosx)/ (1+cosx) `
`= (2sin^2 (x/2))/(2 cos^2(x/2))= tan^2 (x/2)`
Answer
Discussion
27.

Find the principal value of each of the following:(i)`sin^(-1)1/2-2sin^(-1)1/(sqrt(2))``"""""`(ii) `sin^(-1){cos(sin^(-1)(sqrt(3/2))}`

Answer» (i) `sin^-1 (1/2) - 2sin(1/sqrt2) = pi/6 - 2(pi/4) `
`=> pi/6 - pi/2 = -pi/3`
(ii) `sin^-1 {cos (sin^-1 (sqrt3/2))} = sin^-1 (cos (pi/3)) `
`= sin^-1 (1/2) = pi/6`
Answer
Discussion
28.

Solve the following equations:`sin[2cos^(-1)"{"cot"("2tan^(-1)x"}]"=0`

Answer» `sin(2cos^(1-)(cot(2tan^(-1)x)))=0`
`2cos^(-1)(cot(2tan^(-1)x))=sin^(-1)0=0`
`cos^(-1)(cot(2tan^(-1)x))=0`
`cot(2tan^(-1)x)=cos0`
`cot(2tan^(-1)x)=1`
`cot(tan^(-1)((2x)/(1-x^2)))=1`
`(1-x^2)/(2x)=1`
`1-x^2=2x`
`x^2+2x-1=0`
`x=(-2pmsqrt(4+4))/2`
`x=-1pmsqrt2`.
Discussion
29.

Evaluate: `tan(2tan^(-1)(1/5))`

Answer» We have =`tan(2tan^-1 1/5) `
as `[2 tan^-1 x = tan^-1((2x)/(1-x^2))]`
so,`tan(tan^-1 ((2(1/5))/(1- (1/5)^2)))`
`= tan(tan^-1 (2/5 xx 25/24))`
`= 5/12`
Answer
Discussion
30.

Find the value of expression: `sin(2tan^(-1)(1/3))+cos(tan^(-1)2sqrt(2))`

Answer» `2 tan^-1 x = sin^-1 (2x)/(1-x^2) `
& `tan^-1 x = cos^-1 (1/sqrt(1 + x^2))`
now, `sin(2 tan^-1 (1/3)) + cos( tan^-1 2 sqrt2)`
`2 tan^-1 (1/3) = sin^-1 ( ( 2 xx 1/3)/(1 + (1/3)^2)) = sin^-1(2/3 xx 9/10) = sin^-1 3/5`
`tan^-1 ( 2 sqrt3) = cos^-1(1/sqrt(1 + 8))`
`= cos^-1 1/3`
now, equating `3/6 + 1/3 = (9+5)/15 = 14/15`
Answer
Discussion
31.

Show that: `cos(2tan^(-1)(1/7))=sin(4tan^(-1)(1/3))`

Answer» LHS=`cos(2 tan^-1 (1/7))`
`cos(tan^-1((1/7)/(1-49)))`
`= cos(tan^-1(27/48))`
`= cos(cos^-1(24/25)) = 24/25`= LHs
RHS= `sin(4tan^-1(1/3)`
`= sin(2(2tan^-1(1/3)))`
`= sin(2 tan^-1 ((2/3)/(1-1/9)))`
`= sin(2 tan^-1 (3/4))`
`= sin(sin^-1((2 xx3/4)/(1 + 9/16)))`
`= 24/25= `RHS
so, LHS=RHS
hence proved
Discussion
32.

Solve the following equation for `x :``tan^(-1)(1/4)+2tan^(-1)(1/5)+tan^(-1)(1/6)+tan^(-1)(1/x)=pi/4`

Answer» `2 tan^-1 x = tan^-1((2x)/(1-x^2))`
now,`tan^-1 (1/4) + tan^-1((2/5)/(1-(1/25))) + tan^-1(1/6) + tan^-1 (1/x) = pi/4`
=`tan^-1(1/4) + tan^-1(5/12) + tan^-1(1/6) + tan^-1(1/x) = tan^-1 1`
`= tan^-1((1/4 + 5/12)/(1- 5/48)) + tan^-1(1/6) + tan^-1 (1/x) = tan^-1 1`
`= tan^-1(32/43) + tan^-1(1/6) + tan^-1(1/x) = tan^-1(1) `
`= tan^-1((32/43 + 1/6)/(1- (32/43)(1/6))) + tan^-1(1/x) = tan^-1 1`
`= tan^-1(((192+43)/258)/((258- 32)/(258))) + tan^-1 (1/x)= tan^-1 1 `
`tan^-1 (235/226) + tan^-1(1/x) = tan^-1 1`
`tan^-1 (1/x) = tan^-1 1 - tan^-1(235/226)`
`= tan^-1((1- 235/226)/(1 + 235/226))`
`tan^-1 (1/x) = tan^-1((226-235)/(226 + 235))`
`1/x = -9/461`
`x= -461/9`
Answer
Discussion