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151.

सिद्ध करे कि : `tan((pi)/(4) +(1)/(2) cos^(-1) ""(a)/(b)) +tan((pi)/(4) - (1)/(2) cos^(-1)""(a)/(b)) = (2b)/(2)`

Answer» माना कि `cos^(1) ""(a)/(b) = theta " "therefore cos theta = (a)/(b)` अब `L.H.S. = tan((pi)/(4) +(theta)/(2)) +tan((pi)/(4) - (theta)/(2))`
` = (1+tan ""(theta)/(2))/(1-tan ""(theta)/(2)) +(1-tan""(theta)/(2))/(1+tan""(theta)/(2))=((1+tan ""(theta)/(2))^(2) +(1-tan ""(theta)/(2))^(2))/(1-tan^(2)""(theta)/2)`
` = (2(1+tan^(2) ""(theta)/(2)))/(1-tan^(2)""(theta)/(2))=(2)/(cos theta) = (2)/(a//b) = (2b)/(a) =R.H.S.`
Discussion
152.

सिद्ध कीजिए कि `sin^(-1)((2a)/(1+a^(2)))+sin^(-1)((2b)/(1+b^(2)))=2tan^(-1)""(a+b)/(1-ab)`

Answer» बायाँ पक्ष `=2 tan^(-1)a+2 tan^(-1)b`
`= 2[tan^(-1)a+tan^(-)b]`
`=2[tan^(-1)""(a+b)/(a-ab)]=` दायाँ पक्ष
Discussion
153.

यदि `sin^(-1)x+tan^(-1)x=(pi)/(2)` तो सिद्ध कीजिए कि `2x^(2)+1=sqrt5`

Answer» `impliestan^(-1)x=(pi)/(2)-sin^(-1)x=cos^(-1)x=tan^(-1)""(sqrt(1-x^(2)))/(x)`
`impliesx=(sqrt(1-x^(2)))/(x)" implies "x^(2)=sqrt(1-x^(2))`
`x^(4)=1-x^(2)" implies "x^(4)+x^(2)-1=0`
`impliesx^(2)=(-1+-sqrt(1+4))/(2xx1)" implies "2x^(2)=-1+-sqrt5`
`implies2x^(2)+1=sqrt5(":.2x^(2)+1=-sqrt5` सम्भव नहीं है।)
Discussion
154.

सिद्ध कीजिए कि - `tan^(-1)1+tan^(-1)2+tan^(-1)3=pi`

Answer» `LH.S.=tan^(-1)+tan^(-1)2+tan^(-1)3`
`=(pi)/(4)+((pi)/(2)-cot^(-1)2)+((pi)/(2)-cot^(-1)3),`
`[because tan^(-1)x=(pi)/(2)-cot^(-1)x" सभी "x in R " के लिए "]`
`=(pi)/(4)+pi-(cot^(-1)2+cot^(-1)3)`
`=(pi)/(4)+pi-[tan^(-1)((1)/(2))+tan^(-1)((1)/(3))]`
`" "[because cot^(-1)x=tan^(-1)((1)/(x))]`
`=(pi)/(4)+pi-tan^(-1)(((1)/(2)+(1)/(3))/(1-(1)/(2)xx(1)/(3)))`
`=(pi)/(4)+pi-tan^(-1)(((5)/(6))/((5)/(6)))`
`=(pi)/(4)+pi-tan^(-1)(1)`
`=(pi)/(4)+pi-(pi)/(4)`
`=pi=R.H.S." "` यही सिद्ध करना था ।
Discussion
155.

यदि `tan^(-1)x+tan^(-1)y+tan^(-1)z=pi` तो सिद्ध कीजिए कि `x+y+z=xyz`

Answer» `tan^(-1)x+tan^(-1)y+tan^(-1)z=pi`
`impliestan^(-1)""(x+y)/(1-xy)+tan^(-1)z=pi`
`impliestan^(-1)""((x+y)/(1-xy)+z)/(1-(x+y)/(1-xy).z)=pi`
`(x+y+z(1-xy))/(1-xy-(x+y).z)-tanpi=0`
`impliesx+y+z-xyz=0`
`impliesx+y+z=xyz`
Discussion
156.

सरलतम रूप में लिखिए - `tan^(-1){x+sqrt(1+x^(2))}, x in R.`

Answer» Correct Answer - `x=cot theta`
Discussion
157.

किसी त्रिभुज ABC में यदि `A = tan^(-1)2` तथा `B = tan^(-1)3` सिद्ध करे कि `C = (pi)/(4)`.

Answer» दिया है, `A = tan^(-1) 2 rArr tanA = 2` तथा `B = tan^(-1)3 rArr tan B = 3`
चूँकि A,B,C त्रिभुज के कोण है, `therefore A + B + C = pi`
`therefore C = pi - (A+B) " "...(1)`
अब, `A + B = tan^(-1)3 = pi +tan^(-1)[(2+3)/(1-2.3)]`
`[because tan^(-1) x+tan^(-1)y = pi + tan^(-1)[(x+y)/(1-xy)]`
for `x gt 0, y gt 0` तथा `xy gt 1` यहाँ `xy = 2*3 = 6 gt1]`
` = pi + tan^(-1) (-1) = pi - tan^(-1) (1) = pi - (pi)/(4) = (3pi)/(4)`
(1) से, `C = pi - (3pi)/(4)l = (pi)/(4)`
Discussion
158.

यदि `tan^(-1){(sqrt(1+x^(2))-sqrt(1-x^(2)))/(sqrt(1+x^(2))+sqrt(1-x^(2)))}=alpha`, तब सिद्ध कीजिए - `x^(2)=sin 2alpha.`

Answer» `tan^(-1){(sqrt(1+x^(2))-sqrt(1-x^(2)))/(sqrt(1+x^(2))+sqrt(1-x^(2)))}=tan alpha`
`rArr" "=(sqrt(1+x^(2))-sqrt(1-x^(2)))/(sqrt(1+x^(2))+sqrt(1-x^(2)))=tan alpha`
`rArr" "-(sqrt(1+x^(2)))/(sqrt(1-x^(2)))=(tan alpha+1)/(tan alpha-1),`
[योगान्तरानुपात नियम से ]
`rArr" "(sqrt(1-x^(2)))/(sqrt(1+x^(2)))=(1-tanalpha)/(1+tan alpha)`
`rArr" "sqrt((1-x^(2))/(1+x^(2)))=(cos alpha-sinalpha)/(cos alpha+sin alpha)`
`rArr" "(1-x^(2))/(1+x^(2))=((cos alpha-sin alpha)/(cos alpha+sin alpha))^(2)`
`rArr" "(1+x^(2))/(1+x^(2))=(1-sin 2alpha)/(1+sin 2alpha)`
`rArr" "x^(2)=sin 2alpha.`
Discussion
159.

यदि `sin^(-1).(2a)/(1+a^(2))+cos^(-1).(1-b^(2))/(1+b^(2))=2tan^(-1)x,` तब सिद्ध कीजिए - `x=(a+b)/(1-ab).`

Answer» माना `a=tanA` और `b=tanB,` तब `A=tan^(-1)a` और `B=tan^(-1)B`
Discussion
160.

दर्शाइए कि `"sin"^(-1) 3/5 "sin"^(-1)8/17="cos"^(-1)84/85`

Answer» बायां पक्ष `"sin"^(-1)3/5-"sin"^(-1)8/17`
माना `"sin"^(-1)3/5=A` तब `"sin"A=3/5`
और `"sin"^(-1)8/17=B` तब `"sin"B=8/17`
अब `"cos" A =sqrt(1-"sin"^(2)A)=sqrt(1-9/25)`
`=sqrt(16/25)=4/5`
तथा `"cos"B=sqrt(1-"sin"^(2)B)=sqrt(1-64/289)`
`=sqrt(225/289)=15/17`
इस‌िलए `"cos"(A-B)="cos"A"cos"B+"sin"A "sin" B`
`=4/5xx15/17+3/5xx8/17`
`60/85+24/85`
`=84/85`
`A-B="cos"^(-1)84/85`
अत: `"sin"^(-1)3/5-"sin"^(-1)8/17="cos"^(-)84/85`
Discussion
161.

यदि `cos^(-1)""(x)/(a)+cos^(-1)""(y)/(b)=theta` तो सिद्ध कीजिए कि `(x^(2))/(a^(2))-(2xy)/(ab).costheta+(y^(2))/(b^(2))=sin^(2)theta`.

Answer» `cos^(-1)""(x)/(a)+cos^(-1)""(y)/(b)=theta`
`impliescos^(-1)[(x)/(a).(y)/(b)-sqrt(1-(x^(2))/(a^(2)))sqrt(1-(y^(2))/(b^(2)))]=theta`
`implies(xy)/(ab)-sqrt(1-x^(2)/(a^(2))-y^(2)/(b^(2))+(x^(2)y^(2))/(a^(2)b^(2)))=costheta`
`implies(xy)/(ab)costheta=sqrt(1-(x^(2))/(a^(2))-(y^(2))/(b^(2))+(x^(2)y^(2))/(a^(2)b^(2)))`
दोनों पक्षों का वर्ग करने पर
`(x^(2)y^(2))/(a^(2)b^(2))+cos^(2)theta-(2xy)/(ab)costheta=1-(x^(2))/(a^(2))-(y^(2))/(b^(2))+(x^(2)y^(2))/(a^(2)b^(2))`
`implies(x^(2))/(a^(2))-(2xy)/(ab)costheta+(y^(2))/(b^(2))=1-cos^(2)theta`
`implies(x^(2))/(a^(2))-(2xy)/(ab)costheta+(y^(2))/(b^(2))=sin^(2)theta`
Discussion
162.

सिद्ध कीजिए कि `sin^(-1)""(3)/(5)-sin^(-1)""(8)/(17)=cos^(-1)""(84)/(85)`

Answer» माना `sin^(-1)""(3)/(5)=theta_(1)` तथा `sin^(-1)""(8)/(17)=theta_(2)`
तब `sin theta_(1)=(3)/(5)` तथा `sin theta_(2)=(8)/(17)`
अब `cos theta_(1)=sqrt(1-sin^(2)theta_(1))=sqrt(1-(9)/(25))=sqrt((16)/(25))=(4)/(5)`
तथा `cos theta_(2)=sqrt(1-sin^(2)theta_(2))=sqrt(1-(64)/(289))=sqrt((255)/(289))=(15)/(17)`
अब , `cos (theta_(1)-theta_(2))=cos theta_(1) cos theta_(2) + sin theta_(1) sin theta_(2)`
`=((4)/(5)xx(15)/(17))+((3)/(5)xx(8)/(17))`
`=(12)/(17)+(24)/(85)=(84)/(85)`
इसलिए , ` theta_(1)-theta_(2)=cos^(-1)""((84)/(85))`
अतः `sin^(-1)""(3)/(5)-sin^(-1)""(8)/(17)=cos^(-1)""((84)/(85))`
Discussion
163.

दर्शाइये कि ` sin^(-1)""((8)/(17))+sin^(-1)""((3)/(5))=cos^(-1)""((36)/(85))`

Answer» माना
बायाँ पक्ष `=sin^(-1)""((8)/(17))+sin^(-1)""((3)/(5))`
`sin^(-1)""[(8)/(17)sqrt(1-((3)/(5))^(2))+(3)/(5)sqrt(1-((8)/(17))^(2))]`
`sin^(-1)""((8)/(17)xx(4)/(5)+(3)/(5)xx(15)/(17))=sin^(-1)""((77)/(85))`
`=cos^(-1)""(sqrt(1-((77)/(85))^(2)))=cos^(-1)""((36)/(85))=` दायाँ पक्ष
Discussion
164.

सिद्ध कीजिए - `sin^(-1).(3)/(5)+sin^(-1).(8)/(17)=sin^(-1).(77)/(85).`

Answer» L.H.S. `=sin^(-1).(3)/(5)+sin^(-1).(8)/(17)`
`=sin^(-1)[(3)/(5)(sqrt(1-((8)/(17))^(2)))+(8)/(1)sqrt(1-((3)/(5))^(2))]`
`[because sin^(-1)x+sin^(-1)y=sin^(-1)(x sqrt(1-y^(2))+ysqrt(1-x^(2)))]`
`=sin^(-1)[(3)/(5)xxsqrt((289-64)/((17)^(2)))+(8)/(17)sqrt((25-9)/((5)^(2)))]`
`=sin^(-1)((3)/(5)xxsqrt((225)/(289))+(8)/(17)xxsqrt((16)/(25)))`
`=sin^(-1)[(3)/(5)xx(15)/(17)+(8)/(17)xx(4)/(5)]`
`=sin^(-1)((9)/(17)+(32)/(85))`
`=sin^(-1)((77)/(85))`
= R.H.S.
यही सिद्ध करना था।
Discussion
165.

सिद्ध कीजिए कि - `tan{(pi)/(4)+(1)/(2)cos^(-1).(a)/(b)}+tan{(pi)/(4)-(1)/(2)cos^(-1).(a)/(b)}=(2b)/(a).`

Answer» माना `cos^(-1)((a)/(b))=theta,` तब `cos theta=(a)/(b)`
`therefore L.H.S.=tan((pi)/(4)+(theta)/(2))+tan((pi)/(4)-(theta)/(2))`
`=(1+tan.(theta)/(2))/(1-tan.(theta)/(2))+(1-tan.(theta)/(2))/(1-tan.(theta)/(2))+=((1+tan.(theta)/(2))^(2)+(1-tan.(theta)/(2))^(2))/((1-tan^(2).(theta)/(2)))`
`=2((1+tan^(2).(theta)/(2))/(1-tan^(2).(theta)/(2)))=(2)/(cos theta)=(2b)/(a)`
= R.H.S.
यही सिद्ध करना था ।
Discussion
166.

यदि `cos^(-1)x + cos^(-1)y + cos^(-1) z = x`, सिद्ध करे कि `x^(2) + y^(2) + z^(2) + 2xyz = 1.`

Answer» माना कि, `cos^(-1) x = alpha, cos^(-1) y = beta` तथा `cos^(-1) z = gamma `
`rArr cos alpha = x, cos beta = y` तथा `cos gamma = z `
प्रश्न से, `alpha + beta + gamma = pi therefore alpha + beta = pi -gamma`
`therefore cos (alpha + beta) = cos (pi-gamma)`
या, `cos alpha cos beta - sin alpha sin beta = - cos gamma`
या, `xy-sqrt(1-x^(2)) sqrt(1-y^(2)) = -z` या, `xy + z = sqrt(1-x^(2)) sqrt(1-y^(2))`
दोनों तरफ करने पर, `x^(2) y^(2) + z^(2) + 2xyz = 1 - x^(2) - y^(2) + x^(2) y^(2)`
या, `x^(2) + y^(2) + z^(2) + 2xyz = 1`
Discussion
167.

यदि `tan^(-1)a+tan^(-1)b+tan^(-1)c=pi`, तब सिद्ध कीजिए - `a+b+c=abc`.

Answer» माना `tan^(-1) a= x, tan^(-1) b =y`
और `tan^(-1)c = z`, तब
`a=tan x, b = tan y` और `c = tan z`.
यहाँ
`tan^(-1)a+tan^(-1)b+tan^(-1)c=pi`
`rArr " " x+y+z=pi`
`rArr " " x+y=pi-z`
दोनों पक्षों का tangent लेने पर,
`tan(x+y)=tan(pi-z)`
`rArr (tan x + tan y)/(1-tan x tan y) = - tan z`
`rArr " " tan x + tan y = - tan z(1-tan x tan y)`
`rArr tan x + tan y + tan z = tan x tan y tan z`
`rArr " " a+b+c=abc`.
यही सिद्ध करना था |
Discussion
168.

सिद्ध कीजिए - `"sin"^(-1)(3)/(5)-"sin"^(-1)(8)/(17)="cos"^(-1)(84)/(85)`.

Answer» L.H.S. `= "sin"^(-1)(3)/(5)-"sin"^(-1)(8)/(17)`
`=cos^(-1)sqrt(1-((3)/(5))^(2))-cos^(-1)sqrt(1-((8)/(17))^(2))`
`= cos^(-1)sqrt(1-(9)/(25))-cos^(-1)sqrt(1-(64)/(289))`
`= "cos"^(-1)(4)/(5)-"cos"^(-1)(15)/(17)`
`= cos^(-1)[(4)/(5)xx(15)/(17)+sqrt(1-((4)/(5))^(2))sqrt(1-((15)/(17))^(2))]`
`= cos^(-1)[(60)/(85)+(3)/(5)xx(8)/(17)]`
`= cos^(-1)((60)/(85)+(24)/(85))`
`= cos^(-1)((84)/(85))`
= R.H.S.
यही सिद्ध करना था |
Discussion
169.

यदि `"cos"^(-1)(x)/(a)+"cos"^(-1)(y)/(b)=alpha` तब सिद्ध कीजिए - `(x^(2))/(a^(2))-(2xy)/(ab)cos alpha + (y^(2))/(b^(2))=sin^(2)alpha`.

Answer» यहाँ
`"cos"^(-1)(x)/(a)+"cos"^(-1)(y)/(b)=alpha`
`rArr cos^(-1)[(x)/(a).(y)/(b)-sqrt(1-(x^(2))/(a^(2)))sqrt(1-(y^(2))/(b^(2)))]=alpha`
`rArr cos^(-1)[(xy)/(ab)-sqrt(1-(x^(2))/(a^(2)))sqrt(1-(y^(2))/(b^(2)))]=alpha`
`rArr " " (xy)/(ab)-sqrt(1-(x^(2))/(a^(2)))sqrt(1-(y^(2))/(b^(2)))=cos alpha`
`rArr ((xy)/(ab)-cos alpha)= sqrt(1-(x^(2))/(a^(2)))sqrt(1-(y^(2))/(b^(2)))`
दोनों पक्षों का वर्ग करने पर,
`((xy)/(ab)-cos alpha)^(2)= (1-(x^(2))/(a^(2)))(1-(y^(2))/(b^(2)))`
`rArr (x^(2)y^(2))/(a^(2)b^(2))-(2xy)/(ab)cos alpha + cos^(2)alpha`
`= 1-(x^(2))/(a^(2))-(y^(2))/(b^(2))+(x^(2)y^(2))/(a^(2)b^(2))`
`rArr (x^(2))/(a^(2))+(y^(2))/(b^(2))-(2xy)/(ab)cos alpha = 1- cos^(2)alpha`
`rArr (x^(2))/(a^(2))+(y^(2))/(b^(2))-(2xy)/(ab)cos alpha = sin^(2)alpha`.
यही सिद्ध करना था |
Discussion
170.

यदि `cos^(-1)x+cos^(-1)y+cos^(-1)z=pi`, सिद्ध कीजिए कि `x^(2)+y^(2)+z^(2)+2xyz=1`

Answer» माना `cos^(-1)x=A, cos^(-1)y=B , cos^(-1)z=C`
`implies cos A =x , cos B=y,cos C=z`
`:. A+B+C=piimplies A+B=pi-C`
`implies cos(A+B)=cos(pi-C)`
`implies cosA cos B- sinA sin B=-cosC`
`impliescosA cosB-sinA sinB+cosC=0`
`implies cosA cosB + cosC=sinA sin B`
`implies cosA cosB +cosC=sqrt(1-cos^(2)A)sqrt(1-cos^(2)B)`
`implies xy+z=sqrt(1-x^(2))sqrt(1-y^(2))`
`implies (xy)^(2)+z^(2)+2xyz=(1-x^(2))(1-y^(2))` ( दोनों ओर का वर्ग करने पर )
`implies x^(2)y^(2)+z^(2)+2xyz=1-y^(2)-x^(2)+x^(2)y^(2)`
`implies x^(2)+y^(2)+z^(2) +2xyz=1`
Discussion
171.

यदि `cos^(-1)x+cos^(-1)y+cos^(-1) z = pi`, तब सिद्ध कीजिए - `x^(2)+y^(2)+z^(2)+2xyz=1`.

Answer» यहाँ
`cos^(-1)x+cos^(-1)y+cos^(-1)z=pi`
`rArr " " cos^(-1)x + cos^(-1)y =pi - cos^(-1)z`
`rArr cos^(-1)(xy - sqrt(1-x^(2))sqrt(1-y^(2)))=cos^(-2)(-z), " " [because cos^(-1)(-z)=pi-cos^(-1)z]`
`rArr " " xy - sqrt(1-x^(2))sqrt(1-y^(2))= -z`
`rArr " " xy+z=sqrt(1-x^(2))sqrt(1-y^(2))`
दोनों पक्षों का वर्ग करने पर,
`(xy+z)^(2)=(1-x)^(2)(1-y^(2))`
`rArr x^(2)y^(2)+z^(2)+2xyz=1-x^(2)-y^(2)+x^(2)y^(2)`
`rArr " " x^(2)+y^(2)+z^(2)+2xyz=1`.
यही सिद्ध करना था |
Discussion
172.

यदि `sin^(-1)x+sin^(-1)y+sin^(-1)z=pi` तब सिद्ध कीजिए - `xsqrt(1-x^(2))+ysqrt(1-y^(2))+z sqrt(1-z^(2))=2xyz`

Answer» माना `sin^(-1)x=A, sin^(-1)y=B`
और `sin^(-1)z=C,` तब
`sinA=x, sin B=y" और "sinC=z" …(1)"`
यहाँ `sin^(-1)x+sin^(-1)y+sin^(-1)z=pi`
`rArr" "A+B+C=pi" …(2)"`
हम जानते हैं कि ,
`sn2A+sin2B+sin2C=4sin A sin B sin C`
`rArr 2sinA cosA+2sin B cosB+2sin C cos C`
`" "4sinA sin BsinC`
`rArr" "sinA sqrt(1-sin^(2)A)+sinBsqrt(1-sin^(2)B)+sinCsqrt(1-sin^(2)C)=2sinAsinBsinC`
`rArr" "x sqrt(1-x^(2))+y sqrt(1-y^(2)) +z sqrt(1-y^(2))+z sqrt(1-z^(2))=2xyz`
यही सिद्ध करना था ।
Discussion
173.

`sin [ cos^(-1) (-(1)/(2))]` का मान निकालें|

Answer» `= sin""(pi)/(3)-(-(pi)/(3))] = sin[(pi)/(3) + (pi)/(6)] = sin""(pi)/(2) =1`
`= sin[pi - cos^(-1)""(1)/(2)] " "[because cos^(-1)(-(1)/(2)) = pi-cos^(-1)""(1)/(2)]`
`= sin [pi-(pi)/(3) ] = sin ""(2pi)/(3) = (sqrt(3))/(2) " "[because cos^(-1)((1)/(2)) = (pi)/(3)]`
Discussion
174.

`sin [(pi)/(3)-sin^(-1)(-(1)/(2))]` का मान निकालें|

Answer» `sin[(pi)/(3) -sin^(-1)(-(1)/(2))]`
` = sin[(pi)/(3)-{-sin^(-1)((1)/(2))}] [because sin^(-1)(-x) = - sin^(-1)x, ]`
`= sin""(pi)/(3)-(-(pi)/(3))] = sin[(pi)/(3) + (pi)/(6)] = sin""(pi)/(2) =1`
Discussion
175.

`sin^(-1)(-(sqrt(3))/(2))` का ज्ञात करें |

Answer» यहाँ हमे प्रतिलोम त्रिकोणीय फलनों का मुख्य मान (Principal value) लेना होगा |
माना कि `sin^(-1) (-(sqrt(3))/(2)) = theta`
`rArr sin theta = - (sqrt(3))/(2)` तथा `- (pi)/(2) le theta le (pi)/(2)`
`rArr theta = - (pi)/(3) rArr sin^(-1)(-(sqrt(3))/(2)) = - (pi)/(3)`
Discussion
176.

`cot^(-1) (-1)` का ज्ञात करें |

Answer» यहाँ हमे प्रतिलोम त्रिकोणीय फलनों का मुख्य मान (Principal value) लेना होगा |
माना कि `cot^(-1) (-1) = theta`, तो `cot theta = -1` तथा `0 lt theta lt pi`
`rArr theta = (3pi)/(4)`, अर्थात `cot^(-1) = (3pi)/(4)`
Discussion
177.

`tan^(-1)(-1)` का ज्ञात करें |

Answer» यहाँ हमे प्रतिलोम त्रिकोणीय फलनों का मुख्य मान (Principal value) लेना होगा |
माना कि `tan^(-1) (-1) = theta`
तो `tan theta = -1` तथा `-(pi)/(2) lt theta lt (pi)/(2)`
`rArr theta = - (pi)/(4) " " [because -(pi)/(2) lt theta lt (pi)/(2)]`
अतः, `tan^(-1) (-1) = - (pi)/(4)`
Discussion
178.

`tan^(-1) ""(cos x)/(1-sinx )` का सरलतम रूप में लिखें

Answer» `(pi)/(4) + (x)/(2)`
Discussion
179.

हल कीजिए ` sin^(-1)x+sin^(-1)(1-x)=cos^(-1)x `

Answer» Correct Answer - `(1)/(2),0`
Discussion
180.

ज्ञात करें : `tan[(1)/(2) cos^(-1)""(sqrt(5))/(3)]`

Answer» माना कि `cos^(-1)((sqrt(5))/(2))= 2theta` तो `cos 2theta = (sqrt(5))/(3)` तथा `0 le 2theta le pi`
अब, `cos 2theta = (sqrt(5))/(3)`
`therefore (1-tan^(2)theta)/(1+tan^(2)theta) = (sqrt(5))/(3)` या, `(1+tan^(2)theta)/(1-tan^(2)theta) = (3)/(sqrt(5))`
या, `(2tan^(2)theta)/(2) = (3-sqrt(5))/(3+sqrt(5)) " "` [योगान्तर निष्पत्ति से]
या, `tan^(2) = ((3-sqrt(5))/(3+sqrt(5)))((3-sqrt(5))/(3-sqrt(5)))" "...(i)`
लेकिन `0 le 2theta le pi " " therefore le theta le (pi)/(2)`
`rArr theta` पहले पाद में है | अतः `tan theta` धन होगा |
`therefore` (1) से, `tan theta = (3-sqrt(5))/(2)`
Discussion
181.

मान निकालें : `cot^(-1) cot((5pi)/(4))`

Answer» Correct Answer - `(pi)/(4)`
Discussion
182.

निम्न समीकरण को हल कीजिए - `tan^(-1)(x-1)+tan^(-1)(x)+tan^(-1)(x+1)=tan^(-1) 3 x `

Answer» दिया है ` tan^(-1)""(x-1+x+x+1-(x-1)x(x+1))/(1-(x-1)x-x(x+1)-(x-1)(x+1))=tan^(-1)3x`
` tan^(-1)""(3x-x(x^(2)-1))/(1-x^(2)+x-x^(2)-x-x^(2)+1)=tan^(-1)3x`
`(3x-x^(3)+x)/(2-3x^(2))=(3x)/(1)`
`4x-x^(3)=6x-9x^(3)`
`8x^(3)-2x=0`
`2x(4x^(2)-1)=0`
x=0
तथा `4x^(2)-1=0`
` 4x^(2)=1`
`x^(2)=(1)/(4)implies x=pm(1)/(2)`
इसलिए `x=0, pm(1)/(2)`
Discussion
183.

सिद्ध कीजिए कि `"tan"^(-1)5-"tan"^(-1)3+"tan"^(-1)7/9=(pi)/4`

Answer» वाम पक्ष
`="tan"^(-1)5-"tan"^(-1)3+"tan"^(-1)7/9`
`"tan"^(-1)(8xx3)/(1+5xx3)+"tan"^(-1)7/9`
`="tan"^(-1)2/16+"tan"^(-1)7/9`
`="tan"^(-1)(2/16+7/9)/(1-2/16+7/9)`
`="tan"^(-1)((18+12)/144)/((144-14)/144)`
`="tan"^(-1)130/130`
`="tan"^(-1)(1)="tan"^(-1)(""tan""(pi)/4)`
`=(pi)/4=` दायां पक्ष
Discussion
184.

हल कीजिए : `4sin^(-1)x+cos^(-1)x=pi`.

Answer» Correct Answer - (i) `(1)/(2)` (ii) 13
Discussion
185.

निकालें `sin^(-1)(sin ""(2pi)/(3))`

Answer» माना कि `sin^(-1)(sin""(2pi)/(3)) = theta rArr sin theta = sin""(2pi)/(3)` तथा ` - (pi)/(2) le theta le(pi)/(2)`
`rArr sin theta = (sqrt(3))/(2) rArr theta = (pi)/(3)" "[because -(pi)/(2) le theta le (pi)/(2)]`
अतः `sin^(-1)(sin "" (2pi)/(3)) = (pi)/(2)`
Discussion
186.

x के लिए हल करें : `sin (sin^(-1)""(1)/(5) + cos^(-1) x) = 1`

Answer» Correct Answer - `(1)/(5)`
Discussion
187.

सिद्ध कीजिए `"cos"^(-1)((1-a^(2))/(1+a^(2)))-"cos"^(-1)((1-b^(2))/(1+b^(2)))=2"tan"^(-1)((a-b)/(1+ab))`

Answer» बायां पक्ष
`"cos"^(-1)((1-a^(2))/(1+a^(2)))-"cos"^(-1)((1-b^(2))/(1+b^(2)))`
माना `a="tan" alpha` तथा `b="tan"beta` रखने पर
`="cos"^(-1)((1-"tan"^(2)alpha)/(1+"tan"^(2)alpha))-"cos"^(-1)((1-"tan"^(2)beta)/(1+"tan"^(2)beta))`
`="cos"^(-1)("cos" 2 alpha)-"cos"^(-1)("cos" 2beta)`
`=2alpha-2beta`
`=2(alpha-beta)`
`=2("tan"^(-1)a-"tan"^(-1)b)`
`=2"tan"^(-1)((a-b)/(1+ab))=` दायां पक्ष
Discussion
188.

हल कीजिए - ` cos^(-1)x + sin^(-1)""((1)/(2)x)=(pi)/(6)`

Answer» Correct Answer - `x=pm1 " और " x=pm sqrt((3)/(7))`
Discussion
189.

यदि `cos^(-1)""(p)/(a)+cos^(-1)""(q)/(b)=alpha ` , सिद्ध कीजिए कि ` (p^(2))/(a^(2))-(2pq)/(ab)cos alpha+(q^(2))/(b^(2))=sin^(2) alpha`

Answer» मान `cos^(-1)""(p)/(a)=A, cos^(-1)""(q)/(b)=B`
`implies cosA =(p)/(a), cos B=(q)/(b)`
`:. A+B=alpha `
`implies cos (A+B)=cos alpha `
`cosA cos B-sinA sinB= cos alpha `
` cos A * cosB- cos alpha=sinA sinB`
`implies (p)/(a) xx (q)/(b)- cos alpha=sqrt(1-(p^(2))/(a^(2)))sqrt(1-(q^(2))/(b^(2)))`
दोनों ओर का वर्ग करने पर
` ((pq)/(ab))^(2)+cos^(2) alpha-(2pq)/(ab)cos alpha=(1-(p^(2))/(a^(2)))(1-(q^(2))/(b^(2)))`
`implies (p^(2)q^(2))/(a^(2)b^(2))+ cos^(2)alpha-(2pq)/(ab) cos alpha=1-(q^(2))/(b^(2))-(p^(2))/(a^(2))+(p^(2)q^(2))/(a^(2)b^(2))`
`implies (p^(2))/(a^(2))-(2pq)/(ab)cos alpha+(q^(2))/(b^(2))=1-cos^(2)alpha`
`implies (p^(2))/(a^(2))-(2pq)/(ab)cos alpha +(q^(2))/(b^(2))=sin^(2)alpha `
Discussion
190.

हल कीजिए - `tan^(-1)""((1-x)/(1+x))-(1)/(2)tan^(-1)x=0 " " (x gt 0)`

Answer» Correct Answer - `(1)/(sqrt(3))`
Discussion
191.

`tan^(-1)x + tan^(-1)""(2x)/(1-x^(2)) = (pi)/(3)` का हल है -

Answer» Correct Answer - `tan""(pi)/(9)`
Discussion
192.

`cos[tan^(-1)((3)/(4))]` ज्ञात करें

Answer» Correct Answer - `(4)/(5)`
Discussion
193.

निम्न समीकरण को हल कीजिए - `tan^(-1)((x-1)/(x-2))+tan^(-1)((x+1)/(x+2))=(pi)/(4)`

Answer» `tan^(-1)((x-1)/(x-2))+tan^(-1)((x+1)/(x+2))=(pi)/(4)`
`implies tan^(-1)[(((x-1)/(x-2))+((x+1)/(x+2)))/(1-((x-1)/(x-2))((x+1)/(x+2)))]=(pi)/(4)`
`=((x-1)(x+2)+(x+1)(x-2))/((x-2)(x+2)-(x-1)(x+1))=1`
`implies (x^(2)+x-2+x^(2)-x-2)/(x^(2)-4-x^(2)+1)=1`
`implies 2x^(2)-4=-3 `
`implies x^(2)=(1)/(2)impliesx=pm(1)/(sqrt(2))`
Discussion
194.

निम्न समीकरण को हल कीजिए - ` sin^(-1)x+sin^(-1)2x=(pi)/(3)`

Answer» `sin^(-1)x+sin^(-1)2x=sin^(-1)""(sqrt(3))/(2)`
`implies sin^(-1)x-sin^(-1)((sqrt(3))/(2))=-sin^(-1)(2x)`
`implies sin^(-1)[xsqrt(1-(3)/(4))-(sqrt(3))/(4)sqrt(1-x^(2))]=sin^(-1)(2x)`
`implies (x)/(2)-(sqrt(3))/(2)sqrt(1-x^(2))=-2x`
`implies (5x)/(2)=(sqrt(3))/(2)sqrt(1-x^(2))`
`implies 5x=sqrt(3)*sqrt(1-x^(2))implies 25x^(2)=3(1-x^(2))`
`implies 28x^(2)=3impliesx^(2)=(3)/(28)`
`:. x= pm(1)/(2)sqrt((3)/(7))`
परन्तु `x=(-1)/(2)sqrt((3)/(7))` दी गयी समीकरण को संतुष्ट नहीं करता इसलिए `x=(1)/(2)sqrt((3)/(7))` दी गयी समीकरण का हल है ।
Discussion
195.

हल करे : `sin^(-1) x + sin^(-1) ""(sqrt(3))/(2)`

Answer» दिया है, `sin^(-1) x + sin^(-1) 2x = sin^(-1) ""(sqrt(3))/(2)`
`rArr sin^(-1) x - sin^(-1) ""(sqrt(3))/(2) = -sin^(-1) 2x" "...(1)`
`rArr sin^(-1)[x*sqrt(1-(3)/(4)) - (sqrt(3))/(2) sqrt(1-x^(2))] = sin^(-1) [-2x]`
`[because sin^(-1) (-x) = - sin^(-1)x]`
`rArr (x)/(2)-(sqrt(3))/(2) sqrt(1-x^(2)) = -2x`
`rArr 5x = sqrt(3) sqrt(1-x^(2))" "...(2)`
वर्ग करने पर, `25x^(2) = 3(1-x^(2))` या, `28x^(2) = 3 therefore x = pm (sqrt(3))/(2sqrt(7))`
लेकिन `x = - (sqrt(3))/(sqrt(27))` दिए गए समीकरण को संतुष्ट नहीं करता है क्योकि x के क्षण मान के लिए समीकरण (2) का L.H.S. क्षण तथा R.H.S. धन हो जायेगा |
`therefore x = (sqrt(3))/(2sqrt(7)) = (1)/(2) sqrt((3)/(7))`
Discussion
196.

हल करे : `tan^(-1)"" (1-x)/(1+x) = (1)/(2) tan^(-1) x(x gt 0)`

Answer» दिया है, `tan^(-1) x = 2 tan^(-1) ((1-x)/(1+x))`
` = tan^(-1) ""(2((1-x)/(1+x)))/(1-((1-x)/(1+x))^(2))" "[because 2 tan^(-1) x = tan^(-1)""(2x)/(1-x^(2))]`
`= tan^(-1)""(2(1-x)(1+x))/((1+x)^(2)-(1-x)^(2))`
`= tan^(-1) ""(2(1-x^(2)))/(4x)`
`therefore tan^(-1) x = tan^(-1)"" (1-x^(2))/(2x)`
`therefore x = (1-x^(2))/(2x) " "[because tan^(-1)` एकैकी फलन है]
`rArr 2x^(2) = 1-x^(2) " " rArr 3x^(2) = 1`
`rArr x = pm (1)/(sqrt(3)) " " therefore x = (1)/(sqrt(3))" "[because x gt 0]`
Discussion
197.

हल कीजिए `"tan"^(-1)(x+1)+"tan"^(-1)(x-1)="tan"^(-1)8/31`

Answer» `"tan"^(-1)(x+1)+"tan"^(-1)(x-1)="tan"^(-1)8/31`
`"tan"^(-1)(x+1+x-1)/(1-(x+1)(x-1))="tan"^(-1)8/31`
`(2x)/(1-x^(2)+1)=8/31`
`x/(2-x^(2))=4/31`
`31x=8-4x^(2)`
`4x^(2)+31x-8=0`
`4x^(2)+32x-x-8=0`
`4x(x+8)-1(x+8)=0`
`(x+8)(4x-1)=0`
`x=-8, 1/4`
Discussion
198.

हल करें : `tan^(-1) ""(1)/(2) = cot^(-1)x + tan^(-1) ""(1)/(7)`

Answer» Correct Answer - 3
Discussion
199.

हल करें : `tan^(-1) (x -1) + tan^(-1) x + tan^(-1) (x+1) = tan^(-1) 3`

Answer» Correct Answer - `0, pm ""(1)/(2)`
Discussion
200.

हल करें : `tan^(-1) 2x + tan^(-1) 3x = (pi)/(4)`

Answer» Correct Answer - `(1)/(6)`
Discussion