INVERSE TRIGONOMETRY: SA 4 MARKS

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1 INVERSE TRIGONOMETRY: SA MARKS To prove Q. Prove that sin - tan - 7 = π 5 Ans L.H.S = Sin - tan = A- tan - 7 = tan - 7 tan- let A = Sin - 5 Sin A = 5 = tan - ( ( ) ) tan tan A = A = tan- tan tan- = tan- ( x 7 ) = tan - ( 65 ) = 65 tan- () = π RHS Q: Prove that tan - + tan - + tan - = π Ans:- tan - + tan - + tan - = tan - +( tan - + tan - ) = π +π = π = π Q. Prove that sin - + cos- + 6 tan- 6 Ans:- We havesin - + cos - + tan = tan - + tan - + tan = π + tan - ( x ) + tan - 6 = π + tan - ( 6 ) tan- 6 6 = π - tan - ( 6 6 ) + tan- 6 6 = π = π Q. Prove that tan - + tan - 5 tan- 7 tan- 8 = π Ans:- LHS = [tan - + tan - ]+ 5 [tan- ( ) + 7 tan- ( )] 8 = tan - ( ) + tan - ( ) = tan tan- 55 = tan tan- = tan x = tan - ( + ) = tan- 65

2 = tan - () = π = R.H.S Q5. Prove that Sin Sin- 5 + Sin =π Ans: Let A = Sin - 5, and B = Sin- 5 cos A = sin A cos B = sin B = ( 5 ) ( 5 ) = 9 5 = 5 69 = Cos(A+B) = cosa cosb sinasinb = = A+B = cos - ( 6 65 ) = sin- ( 5 )+sin- ( 5 ) = π - sin- ( 6 65 ) [cos- x = π - sin- x)] A+B + sin - ( 6 ) = π 65 Q6. Prove the following : cot - +sinx + sinx +sinx sinx =x,xt(0, π ) Ans: = + sinx = cos x +sin x + sin x cos x = (cos x + sin x ) = cos x + sin x sinx = cos x +sin x sin x cos x = (cos x sin x ) = cos x sin x cos x/ y = cot - ( cosx +sinx +cosx sinx cos x ) = cot - ( ) +sinx cosx +sinx sin x/ =cot- (cot x/) = x/[ x (0, π )] Q7. Prove the following Cos(sin - /5 +cot - / ) = 6 5 Ans:- Let sin - 5 = θ and cot- = = sin θ= 5 and cot = tan θ= / and tan = θ = tan and = = tan Thus sin cot- =tan + tan = tan- [ +. = tan - ( 7/6) = α (say) (i) tan α = 7/6 cos α = Now L.H.S = cos( sin cot- ) = cos α = = 6 5 = RHS ]

3 Q8. Prove the following : Cos - ( 5 ) + Cos - ( ) = Cos - ( 65 ) Ans: LHS Cos - ( 5 ) + Cos - ( ) Cos - ( 5 - ( 5 ) ( ) ) = Cos - ( x 5 ) = cos- ( ) = cos- ( 65 ) = RHS Q9:- Prove the following: Cos - ( ) + sin - ( 5 ) = sin - ( 5 65 ) Ans: L.H.S = Cos - ( ) + sin - ( 5 ) = sin - ( ) + sin - ( 5 ) Sin - ( 5 ) + sin - ( 5 ) = sin - ( 5 ( 5 ) + 5 ( 5 ) = sin - ( 5 x x ) = sin - ( ) = R.H.S Q0. Prove that tan - + tan- 7 + tan- 8 = π Ans: tan - tan- 7 tan- 8 = tan - ( 5 x 5 ) + tan - = tan tan- 8 = tan - ( ) = tan - ( ) = tan- ( ) = tan- () = π Q. Prove that tan - tan- = 7 tan- 7 Ans: = tan - + tan- 7 = tan- ( = tan - ( 7. 7 ) = tan - 7. ( ) tan- = tan - ) 7 tan- 7 Q. Prove that tan - ( +x + x +x x ) = π cos- x L.H.S = tan - ( +x + x +x x ) let x = cos θ

4 LHS = tan - +cos θ + cos θ ( ) = +cos θ cos θ tan- ( cos θ + sin θ ) cos θ sin θ = tan - cos θ +sin θ ( ) cos θ sin θ = +tan θ tan- ( ) tan θ = tan- ( tan π +tan θ tan π ) tan θ = tan - [ tan ( π + θ ) ]= π cos- x = RHS Q. To prove, tan - x = cos - ( x +x ) Let x = tan θ then θ = tan - x RHS = cos - ( x +x ) = cos - ( tan θ +tan θ ) = cos- ( cos θ) = θ = tan - x = LHS Q.To prove sec (tan - ) + cosec ( cot - ) = 5 LHS= sec ( tan - ) + cosec ( cot - ) = + tan (tan - ) + cot ( cot - ) = + [ tan ( tan - )] + [ cot ( cot - )] =+ [] + [] = 5 RHS Q5. To prove, tan ( sin- ¾ ) = 7 Ans: L.H.S = tan( sin- ¾ ) Let sin- ¾ = θ sin - ( ¾) = θ sin θ = ¾ tanθ +tan θ = ¾ 8 tan θ = + tan θ tan θ-8 tan θ + =0 tan θ = 8± 8 6 tan θ = ± 7 = θ = tan - ( ± 7 ) sin- ( ¾) = tan - ( ± 7 ) Taking negative sign, we get sin- ( ¾) = tan - ( 7 ) tan( sin- ¾) = ( 7 ) Q6. Prove that Sin Sin- = 8 7 cos- 85 Ans : we have Sin Sin- = 5 7 cos- 5 -cos- 7 = Cos - { ( 5 ) ( 5 7 ) }

5 = Cos - { } = 5 7 cos- { 60 + } = cos- 85 Q7. Prove that : tan - x + tan - x = x tan- ( x x x), x < Ans:- We have tan - x + tan - x x = tan - { x+ x x x } = tan - ( x x+x x x x ) = tan - ( x x x ) Q8. Prove that Sin [ cot - { cos ( tan - x)}] = x + x + We have, Cos ( tan - x) = cos { cos - +x } = +x sin [ cot - { cos ( tan - x)}] = sin { cot - +x } = sin { +x sin- } = +x +x +x Q9.Prove that Sin - + cos - = 5 cos- = Sin- 65 Ans LHS = Sin - + cos - 5 = Sin - { 5 ( 5 )+ 5 ( 5 ) } = Sin - { + 5 } 5 5 =Sin = cos - ( ) = cos - 65 Q0.Prove the following cos - x = sin - x = cos - x Ans: Let cos - x = y, then x= cos y sin - x = cos y sin- { } = sin - sin y { } = (y) = y And cos - +x = cos - +cos y { } = cos - (cos y/ ) = ( y ) = y Hence cos - x = sin - x = cos- x Q. Prove the following cot - ( xy+ x y ) + cot- ( yz+ y z ) + cot- ( zx+ z x ) =0

6 Ans: tan - ( x y +xy ) + tan- ( y z +yz )+ tan- ( z x +zx )=tan- x- tan - y + tan - y- tan - z+ tan - z- tan - x =0 Q.Solve for x : sin - ( x +x )- cos- ( x +x ) + tan- ( x +x ) =π Ans: Put x= tanθ then x = Q. Solve for x, Sin - x + Sin - x = π Ans: we have Sin - x + Sin - x = π Sin - x = π - Sin- x x = sin ( π - Sin- x) x = sin π cos (Sin- x) cos π sin (Sin- x) x = cos ( cos- x -/ x x= x -/ x x+ x = x on squaring both side we get 5 x = ( -x ) = x = ± 8 Since - does not satisfy the given condition x = 8 8 Q. Tan - (x+) + tan - ( x-) = tan - 8 Ans: tan - [ x = 8 (x ) - x + = x x + x -8 = 0 x++x ] = 8 (x+)(x ) tan- ( x-) ( x+ 8) = 0 x = ¼, x = -8 (rejected) Q5. Solve for x, tan - ( x x ) + tan- ( x+ x+ ) = π Ans: tan - ( x x ) + tan- ( x+ x+ ) = π tan - [ x x +x+ x+ ( x x )(x+ x+ ) ] = π (x )(x+)+(x+)(x ) (x ) (x ) = tan π =

7 x +x +x x = x - = - x = x = ± Q6. Solve for x, tan - (cos x) = tan - (cosec x) cos x tan - ( ) = cos x tan- ( cosec x) cos x = sin x sin x cot x = x= π Q7. Solve for x : cos - ( x x + ) + tan- ( x x ) = π Ans We write cos - {- ( x x + ) }+ tan- {-( x x )} =π π tan - x tan - x = π π = tan- x x = tan π tan ( π -π 6 ) = tanπ 6 +tan π 6 = = - + Q8. Solve for x, tan - (sin x) = tan - ( sec x), x π tan - ( sinx sin x ) = tan- ( cos x ) sinx = cos x cos x sin x cos x = tan x = x = π Q9. Solve the following equation : cos ( tan - x) = sin (cot - ¾) = sec (tan x) cosec ( cot /) Sec( tan - x) = cosec (cot - ¾) + tan x = + cot (cot ) + x = + 9 /6 x = ± Q0. Solve cot - x cot - ( x+) = π Ans. tan - x tan- ( x+ ) = π tan - ( x x+ x ˣ x+ ) = π x=, - (+ )

8 Q. Solve tan - ( x+ x )+ tan- ( x x )= tan- (-7) tan - [ x+ x +x x x+ x ˣ x x ]= tan - (-7) x +x+x x+ x x x + = -7 x x + = 0 (x-) = 0 x= Q. Solve tan - [ +x x +x + x ] = β Ans: We have π - cos- (x ) = π - β π cos- (x ) = β x =± sin Q. Solve tan - x + cot - x = π x x Ans x = Q. Sin - α α + Sin - β β = tan - x Ans tan - α + tan - β = tan - x tan - α+ β ( ) = αβ tan- x α+ β x = ( ). αβ Q5. Solve sin - x + sin - x = π sin - (x) = π - sin- x x = sin ( π - sin- x ) sin π cos(sin- x ) - cos π sin (sin- x ) x = x - x 5x = x 5 x = ( - x ) 8 x = x = 7 (x>0) Q6.Solve the following equation: sin - [x ( x) + (-x) x ] = cos - x Ans x x x +(-x) x = sin ( cos - x) = ( x) x x x + ( x) (-x-) =0 x [ x x - ( x) ] =0 x =0 or x- x = -x x = ½

9 Q7. Solve sin - (-x )- sin - x = π Ans (-x) = sin ( π + sin- x) = cos ( sin - x) = - sin ( sin - x) -x = -x x -x =0 x(x-) =0 x=0,/ x=0,/ x=0, satisfies Q8. Find the solution of the equation tan - x- cot - x = tan - ( ) Ans: tan - x + cot - x = π tan - x = π x = tanπ = Q9. tan - x + x x+ tan- = x+ tan- 6 Ans : tan - { x x+ +x x+ x x+. x x+ } = tan - 6, if (x )(x ) (x+)(x+) < tan - ( x x ) = tan- 6 and 6x (x+)(x+) < 0 x x - =0 and x (x+)(x+) >0 ( x-)(8x+)=0 and x+ (-,-/) (0, ) x = / Q0. Solve tan - x + tan - x = π tan - { x+x 6x } = tan-, if 6x < 6x + 5x- =0 and x < 6 x = -, and 6 -<x< 6 6 Q Simplify the following Cos - ( cos x 5 + sin x) 5 Ans: Let = rcos 5, = r sin θ 5 r= ( 5 ) + ( 5 ) ) = and tan θ = r=,θ = tan - = cos 5, = sin θ 5 r sin θ rcos θ = cos - (cos θcos x + sin θ sinx ) = cos - { cos ( x- θ)} = x- tan -

10 Q. Simplify Sin - ( sinx+cos x ), - π < x <π Ans = Sin - ( sinx cos x ) = Sin - ( sin x cos π + cos x sin π ) = Sin - { sin ( x+ π )} = x+π Q. Simplify following tan - acos θ bsin θ ( ) bcos θ+asin θ a tan x b Ans: tan - ( + a ) = a b tan x tan- b tan- ( tan x) = tan - a b Q.Simplify tan - ( a x x a ax), <x< a tan - { (x a ) (x a ) ( x a ) } = tan- x a Q5. Simplify cos x Tan - ( ), π < x +sin x <π Ans: We have cos x tan - ( ) = +sin x tan- { cos x sinx cos x +sinx + sinx cosx } = tan - { (cosx sinx )(cosx +sinx ) } = tan - { cos x sinx (cos x +sinx ) cos x +sinx } =tan - { tanx +tan x } = tan - { tan ( π - x )} =π - x Q6. Simplify tan - cos x sin x ( ), π < x cos x+ sin x <π Ans: tan - cos x sin x ( ) tan x cos x+ sin x tan- ( ) = +tanx tan- ( tan( π -x)) = π - x Q7. Write the following function in simplest form tan - { a x a+x }, a <x<a putting x = a cos θ

11 tan - { a x a+x } = a a cos θ tan- = a+a cos θ tan- cos θ + cos θ = tan - sinθ = tan - (tan θ ) = θ = x cos θ cos- a Q8. Find the value of the following tan [ sin- x +x + cos- y +y ], x <, y >0 and xy< =tan [ tan A sin- ( ) y +tan A cos- +y ] = tan [ tan A sin- ( ) +tan A cos- ( tan B )] +tan B = tan [ sin- (sin A)+cos - (cos B)] =tan tana+tanb ( A+ B) = tan (A+B) = tan A tanb Q9. Simplify Sin - ( x x + x x ) Ans: Let sin - x = θ and sin - x = = x+y xy x =sin θ and x= n, 0 θ, π sin - (x x + x x ) = sin - (sin θcos + sin cosθ) = sin - ( sinθ+ ) = θ+ = sin - x + sin - x