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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 201. |
ज्ञात करें `sin[(pi)/(2) - sin^(-1)(-(sqrt(3))/(2))]` |
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Answer» माना कि `sin^(-1)(-(sqrt(3))/(2)) = theta,` तो `theta in[(-pi)/(2),(pi)/(2)]` साथ ही `sin theta = (-sqrt(3))/(2) = - sin""(pi)/(3) = sin(-(pi)/(3)) rArr = - (pi)/(3) rArr sin ^(-1)((-sqrt(3))/(2)) = - (pi)/(3)` अब `sin[(pi)/(2) - sin^(-1)((-sqrt(3))/(2))] = sin[(pi)/(2) - ((-pi)/(3))] = sin ""(5pi)/(6) = (1)/(2)` |
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| 202. |
निम्न समीकरण को हल कीजिए `tan^(-1)(x+1)+tan^(-1)(x-1)=tan^(-1)""(8)/(31)` |
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Answer» Correct Answer - `x=(1)/(4),-8` `tan^(-1)(x+1)+tan^(-1)(x-1)=tan^(-1)""(8)/(31)` ` implies tan^(-1)[((x+1)+(x-1))/(1-(x+1)(x-1))]=tan^(-1)""(8)/(31)` `implies (2x)/(1-(x^(2)-1))=(8)/(31)` सरल करे । |
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| 203. |
सरलतम रूप में लिखिए - `tan^(-1).(sqrt(1+x^(2))-1)/(x), x ne 0` |
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Answer» माना `x=tan theta, " तब "theta=tan^(-1)x, theta in (-(pi)/(2),(pi)/(2))` `therefore" "tan^(-1).(sqrt(1+x^(2))-1)/(x)` `=tan^(-1)[(sqrt(1+tan^(2)theta)-1)/(tan^(2)theta)]` `=tan^(-1)[(sectheta-1)/(tan theta)]," "[because sec theta gt 0, (-(pi)/(2),(pi)/(2))"में "]` `=tan^(-1)[(1-cos theta)/(sin theta)]` `=tan^(-1)[(1-(1-sin^(2).(theta)/(2)))/(2sin.(theta)/(2).cos.(theta)/(2))]` `=tan^(-1)((sin.(theta)/(2))/(cos.(theta)/(2)))` `=tan^(-1)(tan.(theta)/(2))` `[because -(pi)/(2)lt theta lt (pi)/(2) rArr -(pi)/(4)lt(theta)/(2)lt(pi)/(4)]` `=(theta)/(2)=(1)/(2)tan^(-1)x`जो कि अभीष्ट सरलतम रूप है। |
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| 204. |
`cos[cos^(-1) ((sqrt(3))/(2))+(pi)/(6)]` ज्ञात करें |
| Answer» Correct Answer - `(1)/(2)` | |
| 205. |
`cot(tan^(-1)a + cot^(-1)a)` में से प्रत्येक का मान निकालें | |
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Answer» `cot(tan^(-1) a + cot^(-1)a)` ` = cot((pi)/(2)) = 0 " "[becausetan^(-1)x+cot^(-1)x = (pi)/(2)]` |
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| 206. |
सरलतम रूप में लिखिए - `tan^(-1).(x)/(sqrt(a^(2)-x^(2))),|x| lt a` |
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Answer» माना `x=a sin theta` तब `theta=sin^(-1).(x)/(a)` और `-(pi)/(2)lt theta lt (pi)/(2)" "[because |x| lt a]` `therefore tan^(-1).(x)/(sqrt(a^(2)-x^(2)))` `=tan^(-1).((a sin theta)/(sqrt(a^(2)-a^(2)sin^(2)theta)))` `=tan^(-1)((a sin theta)/(a cos theta)),` `[because -(pi)/(2) lt theta lt (pi)/(2) rArr cos theta gt 0]` `=tan^(-1)(tan theta)` `=theta` `=sin^(-1).(x)/(a)`जो कि अभीष्ट सरलतम रूप है। |
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| 207. |
यदि `tan^(-1)(1+x)+tan^(-1)(1-x)=(pi)/(6)` तो सिद्ध कीजिए कि `x^(2)=2sqrt(3)` |
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Answer» `tan^(-1)(x+1)+tan^(-1)(x-1)=(pi)/(6)` ` implies tan^(-1)[((x+1)+(x-1))/(1-(x+1)(x-1))]=""(pi)/(6)` `implies (2x)/(1-(1-x^(2)))=tan""(pi)/(6)` `(2)/(x^(2))=(1)/(sqrt(3))` `implies x^(2)=2sqrt(3)` |
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| 208. |
`sin(sin^(-1)x+cos^(-1)x)` में से प्रत्येक का मान निकालें | |
| Answer» `sin(sin^(-1)x +cos^(-1)x) = sin((pi)/(2))=1 " "[because sin^(-1) x+cos^(-1)x=(pi)/(2)]` | |
| 209. |
यदि `tan^(-1)x +tan^(-1)y=(pi)/(2)` तो सिद्ध कीजिए कि `xy=1 ` |
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Answer» `tan^(-1)x+tan^(-1)y=(pi)/(2)` `implies tan^(-1)""[(1+y)/(1-xy)]=(pi)/(2)` `implies (x+y)/(1-xy)=tan""(pi)/(2)=oo=(1)/(0)` `implies 1-xy=0` `implies xy=1 ` |
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| 210. |
मान निकालें : `arc sin""(1)/(2) +arc cos ""(1)/(2)` |
| Answer» Correct Answer - `(pi)/(2)` | |
| 211. |
मान निकालें `tan[(1)/(2)(sin^(-1)""(2x)/(1+x^(2))+cos^(-1)""(1-y^(2))/(1+y^(2)))]` |
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Answer» हम जानते है कि `sin^(-1)"" (2x)/(1+x^(2))=2tan^(-1)x` तथा `cos^(-1)""(1-y^(2))/(1+y^(2))=2tan^(-1)y` अब `L.H.S. = tan(tan^(-1)x+tan^(-1)y)` ` = tan[tan^(-1)((x+y)/(1-xy))] =(x+y)/(1-xy) " "[because tan(tan^(-1)x) =x]` |
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| 212. |
सिद्ध कीजिए - `cos^(-1)x+cos^(-1){(x)/(2)+(sqrt(3-3x^(2)))/(2)}=(pi)/(3).` |
| Answer» माना `cos^(-1)x=alpha rArr x = cos alpha` | |
| 213. |
सिद्ध कीजिए कि ` tan^(-1)""((1)/(2))+tan^(-1)""((1)/(3))=(pi)/(4)` |
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Answer» बायाँ पक्ष `=tan^(-1)""((1)/(2))+tan^(-1)""(1)/(3)` `=tan^(-1)[((1)/(2)+(1)/(3))/(1-(1)/(2)xx(1)/(3))]=tan^(-1)""[((3+2)/(6))/((6-1)/(6))]=tan^(-1)1 ` `=(pi)/(4)=` दायाँ पक्ष |
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| 214. |
हल कीजिए - `2 tan^(-1)(cos x )=tan^(-1)(2 "cosec" x)` |
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Answer» Correct Answer - `x=(pi)/(4)` `2 tan^(-1)(cosx)=tan^(-1)(2"cosec"x)` `implies tan^(-1)""((2cosx)/(1-cos^(2)x))=tan^(-1)(2 "cosec" x)` `implies (2 cosx )/(sin^(2)x)=2 " cosec" x ` `implies cos x=sinx implies tanx=1implies x=(pi)/(4)` |
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| 215. |
`2arc sin""(1)/(2) +3arc tan(-1) +arc cos (-(1)/(2))` ज्ञात करें |
| Answer» Correct Answer - `(11pi)/(12)` | |
| 216. |
सरलतम रूप में लिखिए - `tan^(-1)((1)/(sqrt(x^(2)-1))),|x|gt1` |
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Answer» माना `x="cosec"theta,` तब `theta="cosec"^(-1)x` `therefore" "tan^(-1)((1)/(sqrt(x^(2)-1)))` `=tan^(-1)((1)/(sqrt("cosec"^(2)theta-1)))` `=tan^(-1)((1)/(cot theta))` `=tan^(-1)(tan theta)` `=theta="cosec"^(-1)x` जो कि अभीष्ट सरलतम रूप है। |
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| 217. |
सिद्ध कीजिए ` cos^(-1){cos""(-(pi)/(3))} ne -(pi)/(3)` | इसका मान क्या है ? |
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Answer» Correct Answer - `(pi)/(3)` ` cos^(-1)[cos(-(pi)/(3))]=cos^(-1)[cos""(pi)/(3)]=(pi)/(3)` |
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| 218. |
x के लिए हल कजिए - `cos(tan^(-1)x)=sin(cot^(-1)""(3)/(4))` |
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Answer» माना ` cot^(-1)""(3)/(4)=theta implies cot theta=(3)/(4)` इसलिए `sin theta=(4)/(5)implies theta=sin^(-1)""(4)/(5)` माना `tan^(-1)x=phiimplies tan phi=x ` इसलिए ` cos phi=(1)/(sqrt(1+x^(2)))` अतः `sin(cot^(-1)""(3)/(4))=sin theta= sin(sin^(-1)""(4)/(5))=(4)/(5)` तथा `cos(tan^(-1)x)=cos phi=(1)/(sqrt(1+x^(2)))` इसलिए `(1)/(sqrt(1+x^(2)))=(4)/(5)` `implies (1)/(1+x^(2))=(16)/(25)implies x=pm""(3)/(4)` अतः `x=pm(3)/(4)` |
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| 219. |
`tan^(-1) ""(1)/(sqrt(x^(2)-1)),|x| gt1` का सरलतम रूप में लिखें |
| Answer» `(pi)/(2)-sec^(-1)x` | |
| 220. |
x के लिए हल कजिए - `2 tan^(-1)(sinx)=tan^(-1)(2 sec x ), x ne (pi)/(2)` |
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Answer» माना `2 tan^(-1)(sinx)=tan^(-1)(2sec x)` `implies tan^(-1)""((2 sinx )/(1-sin^(2)x))=tan^(-1)(2 sec x)` `implies tan^(-1)(2 secx tanx)=tan^(-1)(2 sec x )` `implies 2 sec x tan x =2 sec x ` `implies tanx=1=tan""(pi)/(4)` अतः `x=(pi)/(4)` |
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| 221. |
सिद्ध करें कि : `tan^(-1)""(1)/(4) + tan^(-1)""(2)/(9)=(1)/(2) cos^(-1)((3)/(5))` |
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Answer» (1) में हम साबित कर चुके है कि `tan^(-1)""(1)/(4) tan^(-1) ""(2)/(9) =tan^(-1)""(1)/(2) therefore L.H.S. = tan^(-1)""(1)/(2) = (1)/(2) *tan^(-1)""(1)/(2)` ` = (1)/(2) cos^(-1) ((1-(1)/(4))/(1 +(1)/(4)))[because 2tan^(-1) x = c os^(-1)(1-x^(2))/(1+x^(2)),` यहाँ `x = (1)/(2)` ` = (1)/(2) cos^(-1) ((3)/(5)) = R.H.S.` |
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| 222. |
सिद्ध करें कि: `tan^(-1)sqrt(x) = (1)/(2) cos^(-1)((1-x)/(1+x)), x in[0, 1]` |
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Answer» `tan^(-1)sqrt(x) = (1)/(2)(2 tan^(-1)sqrt(x))` ` = (1)/(2) cos^(-1)((1-x)/(1+x))[because tan^(-1) x = cos^(-1)""(1-x^(2))/(1+x^(2)),xge0]` |
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| 223. |
सिद्ध करें कि: `tan^(-1)""(1)/(4)+tan^(-1)""(2)/(9)=(1)/(2)tan^(-1)""(4)/(3)` |
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Answer» `L.H.S. = tan^(-1)((1)/(4))+tan^(-1)((2)/(9))` ` = tan^(-1)[((1)/(4)+(2)/(9))/(1-((1)/(4))((2)/(9)))] [because tan^(-1)x + tan^(-1) y = tan^(-1)""(x+y)/(1-xy)]` `= tan^(-1)[((9+8)/(36))/((36-2)/(36))]=tan^(-1)((17)/(34)) = tan^(-1)((1)/(2))" "...(1)` ` = (1)/(2) tan^(-1) ""(2((1)/(2)))/(1-((1)/(2))^(2))" "[because tan^(-1) x = (1)/(2) (2x)/(1-x^(2))]` ` = (1)/(2) tan^(-1) ""(1)/(1-(1)/(4)) = (1)/(2) tan^(-1) ""(1)/(3//4) = (1)/(2) tan^(-1) ""(4)/(3) = R.H.S.` |
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| 224. |
`tan^(-1) ""(sqrt(1+x^(2))-1)/(x), x ne0` का सरलतम रूप में लिखें |
| Answer» Correct Answer - `(1)/(2) tan^(-1)` | |
| 225. |
प्रश्न संख्या 16 से 18में दिए प्रत्येक व्यंजन का मान ज्ञात कीजिए : `tan(sin^(-1)""(3)/(5)+cos^(-1)""(3)/(2))` |
| Answer» Correct Answer - `(17)/(6)` | |
| 226. |
सिद्ध कीजिए कि ` cos^(-1)x+cos^(-1)""((x)/(2)+(1)/(2)sqrt(3-3x^(2)))=(pi)/(3)` |
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Answer» माना `cos^(-1)x=thetaimplies x=cos theta ` तथा ` sqrt(1-x^(2))=sqrt(1-cos^(2)theta)=sin theta ` अब , माना बायाँ पक्ष `= cos^(-1)x+cos^(-1)((x)/(2)+(sqrt(3))/(2)sqrt(1-x^(2)))` `=theta+cos^(-1)(cos""(pi)/(3)*cos theta+sin""(pi)/(3)*sin theta)` `= theta + cos^(-1)[cos""((pi)/(3)-theta)]=theta+((pi)/(3)-theta)` `= (pi)/(3)=` दायाँ पक्ष |
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| 227. |
सिद्ध करें कि `tan^(-1)[(sqrt(1+x^(2))+sqrt(1-x^(2)))/(sqrt(1+x^(2))-sqrt(1-x^(2)))] = (pi)/(4)+(1)/(2) cos^(-1) x^(2)` |
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Answer» `L.H.S. = tan^(-1)((1+sqrt((1-x^(2))/(1+x^(2))))/(1-sqrt((1-x^(2))/(1+x^(2)))))" "...(1)` `x^(2) = cos 2theta`, रखने पर, `sqrt((1-x^(2))/(1+x^(2)))=sqrt((1-cos2theta)/(1+cos 2theta))` ` = sqrt((2 sin^(2)theta)/(2 cos^(2)theta))=|tan theta|= tantheta[because 0 lesqrt((1-x^(2))/(1+x^(2)))le1therefore 0 letheta le(pi)/(4)]` अब,(1) से. `L.H.S. = tan^(-1)((1+tantheta)/(1-tantheta))` ` = tan^(-1)[tan((pi)/(4)+theta)] =(pi)/(4)+theta=(1)/(2) cos^(-1) x^(2)` `[because cos 2theta = x^(2) therefore2 theta = cos^(-1) x^(2)therefore theta = (1)/(2) cos^(-1) x^(2)]` |
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| 228. |
`tan^(-1)2x+tan^(-1)3x=(pi)/(4)` तो x का मान है -A. 1B. -1C. `(-1)/(6)`D. `(1)/(6)` |
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Answer» Correct Answer - D |
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| 229. |
प्रश्न संख्या 16 से 18में दिए प्रत्येक व्यंजन का मान ज्ञात कीजिए : `sin^(-1)(sin""(2pi)/(3))` |
| Answer» Correct Answer - `(pi)/(3)` | |
| 230. |
`sin^(-1)(sin""(2pi)/(3))` का मान है-A. `(pi)/(3)`B. `(pi)/(4)`C. `(2pi)/(3)`D. इनमे से कोई नहीं। |
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Answer» Correct Answer - A |
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| 231. |
सिद्ध कीजिए कि ` tan^(-1)((sqrt(1+x^(2))+sqrt(1-x^(2)))/(sqrt(1+x^(2))-sqrt(1-x^(2))))=(pi)/(4)+(1)/(2)cos^(-1)x^(2)` |
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Answer» हमें सिद्ध करना है कि ` tan^(-1)((sqrt(1+x^(2))+sqrt(1-x^(2)))/(sqrt(1+x^(2))-sqrt(1-x^(2))))=(pi)/(4)+(1)/(2)cos^(-1)x^(2)` माना `(1)/(2) cos^(-1)x^(2)=thetaimplies x^(2)=cos2 theta` `:. sqrt(1+x^(2))=sqrt(1+cos2 theta)` `=sqrt(2cos^(2)theta)=sqrt(2)cos theta` तथा ` sqrt(10-x^(2))=sqrt(1-cos2 theta)=sqrt(2 sin^(2)theta)=sqrt(2)sin theta ` अब , `(sqrt(1+x^(2))+sqrt(1+x^(2)))/(sqrt(1+x^(2))-sqrt(1-x^(2)))=(sqrt(2)cos theta+sqrt(2)sin theta )/(sqrt(2) cos theta -sqrt(2)sin theta)` `=( cos theta + sin theta )/( cos theta - sin theta )=(1+ tan theta )/(1- tan theta )=(tan""(pi)/(4)+tan theta )/(1-tan""(pi)/(4)tan theta )` `(sqrt(1+x^(2))+sqrt(1-x^(2)))/(sqrt(1+x^(2))-sqrt(1-x^(2)))=tan((pi)/(4)+theta)` `implies tan^(-1)[(sqrt(1+x^(2))+sqrt(1-x^(2)))/(sqrt(1+x^(2))-sqrt(1-x^(2)))]=tan""(pi)/(4)+theta` `implies tan^(-1)[(sqrt(1+x^(2))+sqrt(1-x^(2)))/(sqrt(1+x^(2))-sqrt(1-x^(2)))]=tan""(pi)/(4)+(1)/(2)cos^(-1)x^(2)` |
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| 232. |
सिद्ध कीजिए कि `2 tan^(-1)""(1)/(x)=sin^(-1)""((2x)/(x^(2)+1))` |
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Answer» माना `tan^(-1)""(1)/(x)=thetaimplies (1)/(x)=tan theta implies cot theta =x` `:.` बायाँ पक्ष `=2tan^(-1)""(1)/(x)=2 theta ` अब , दायाँ पक्ष `=sin^(-1)((2x)/(x^(2)+1))` `=sin^(-1)((2cot theta)/(1+cot^(2)theta))=sin^(-1)((2 tan theta )/(1+tan^(2)theta))` `= sin^(-1)(sin2 theta)=2 theta=2tan^(-1)""(1)/(x)` `implies sin^(-1)((2x)/(x^(2)+1))=2tan^(-1)""(1)/(x)` |
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| 233. |
`cot^(-1)(1/(sqrt(x^(2)-1))),xgt1` को सरलतम रूप में लिखिए। |
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Answer» `cot^(-1)(1/(sqrt(x^(2)-1)))` माना `x=sectheta` है तब `theta=sec^(-1)x` `=cot^(-1)(1/(sqrt(sec^(2)theta-1)))` `=cot^(-1)(1/("tan" theta))` `=cot^(-1) (cot theta)` `=theta = sec^(-1)x` |
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| 234. |
यदि `x_(1),x_(2),x_(3),x_(4)` समीकरण `x^(4)-x^(3)sin2prop+x^(2)cos2prop` `-xcosprop-sinprop=0` के मूल है तो `tan^(-1)x_(1)+tan^(-1)x_(2)+tan^(-1)x_(3)+tan^(-1)x_(4)` का मान है-A. `prop`B. `90^(@)-prop`C. `180^(@)-prop`D. `270^(@)-prop` |
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Answer» Correct Answer - B |
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| 235. |
मान ज्ञात कीजिए - `sin^(-1)[sin""((2pi)/(3))]` |
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Answer» Correct Answer - `(pi)/(3)` माना `y=sin^(-1)[sin""((2pi)/(3))]` ` implies siny=sin""((2pi)/(3))=sin(pi-(2pi)/(3))=sin""(pi)/(3)implies y=(pi)/(3)` |
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| 236. |
सिद्ध कीजिए `"tan"^(-1)1+"tan"^(-1)1/2+"tan"^(-1)1/3=(pi)/2` |
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Answer» बायां पक्ष`="tan"^(-1)1+"tan"^(-1) 1/2+"tan"^(-1) 1/3` `="tan"^(-1)(1+1/2)/(1-1 1/2)+"tan"^(-1)1/3` `="tan"^(-1)(3/2)/(1/2)+"tan"^(-1)1/3="tan"^(-1)3+"tan"^(-1)1/3="tan"^(-1)(3+1/3)/(1-3 1/3)` `="tan"^(-1) (10/3)/(1-1)="tan"^(-1)(10/3)/0="tan"^(-1) oo=90^(@)=(pi)/2` `=` दायां पक्ष। |
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| 237. |
मान ज्ञात कीजिए - `tan^(-1)[tan""(-(pi)/(4))]` |
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Answer» Correct Answer - `(pi)/(4)` माना `y=tan^(-1)[tan(-(3pi)/(4))]` ` implies tany=tan""((3pi)/(4))=-tan(pi-(pi)/(4))=-(-tan""(pi)/(4))` `implies tany=tan""(pi)/(4)` अतः `y=(pi)/(4)` |
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| 238. |
सिद्ध कीजिए कि `"cot"^(-1)(ab+1)/(a-b)+"cot"^(-1)(bc+1)/(b-c)+"cot"^(-1)(ca+1)/(c-a)=0` |
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Answer» बांया पक्ष `="cot"^(-1)(ab+1)/(a-b)+"cot"^(-1)(bc+1)/(b-c)+"cot"^(-1)(ca+1)/(c-a)` `="tan"^(-1)(a-b)/(1+ab)+"tan"^(-1)(b-c)/(1+bc)+"tan"^(-1)(c-a)/(1+ca)` `="tan"^(-1)a-"tan"^(-1)b+"tan"^(-1)b-"tan"^(-1)c+"tan"^(-1)c-"tan"^(-1)a` `=0` `=` दायां पक्ष। |
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| 239. |
सरलतम रूप व्यक्त कीजिए - `tan^(-1)(sqrt((1-cosx)/(1+cosx))), lt pi.` |
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Answer» यहाँ `tan^(-1)sqrt((1-cosx)/(1+cosx))` `=tan^(-1).sqrt((2sin^(2).(x)/(2))/(2cos^(2).(x)/(2)))` `=tan^(-1)|tan.(x)/(2)|` `=tan^(-1)(tan.(x)/(2))," "[because 0 le x lt pi rArr 0 le (x)/(2) lt (pi)/(2)]` `=(x)/(2).` जो कि अभीष्ट सरलतम रूप है। |
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| 240. |
सिद्ध कीजिए - `tan^(-1).(2)/(11)+tan^(-1).(7)/(24)=tan^(-1).(1)/(2).` |
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Answer» यहाँ `x=(2)/(11)` और `y=(7)/(24),` तब `xy=(14)/(264)lt 1` `L.H.S.=tan^(-1).(2)/(11)+tan^(-1).(7)/(24)` `=tan^(-1)[((2)/(11)+(7)/(24))/(1-(2)/(11)xx(7)/(24))]` `[because tan6(-1)x+tan^(-1)y=tan^(-1)((x+y)/(1-xy))," यदि "xy lt 1]` `=tan^(-1)(((48+77)/(264))/((264-14)/(264)))` `=tan^(-1)((125)/(250))` `=tan^(-1)((125)/(250))` `=tan^(-1)((1)/(2))`यही सिद्ध करना था। |
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| 241. |
`sin(cot^(-1)x)` का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `(1)/(sqrt(1+x^(2)))` | |
| 242. |
सिद्ध कीजिए - `tan^(-1).(a-b)/(1+ab)+tan^(-1).(b-c)/(1+bc)+tan^(-1).(c-a)/(1+ca)=0.` |
| Answer» `tan^(-1)((x-y)/(1+xy))=tan^(-1)x-tan^(-1)y` | |
| 243. |
मान ज्ञात कीजिए - `sin[(pi)/(3)-sin^(-1)""(-(1)/(2))]` |
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Answer» Correct Answer - 1 `sin[(pi)/(3)-sin^(-1)""(-(1)/(2))]=sin[(pi)/(3)-sin^(-1){sin(-(pi)/(6))}]` `=sin[(pi)/(3)-(-(pi)/(6))]` `=sin""((pi)/(3)+(pi)/(6))=sin""(3pi)/(6)=sin""(pi)/(2)=1 ` |
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| 244. |
निम्नलिखित फलनों को सरलतम रूप में लिखिए : `tan^(-1)""((cos x-sinx)/(cosx+sinx)),(pi)/(4) lt x lt (3pi)/(4)` |
| Answer» Correct Answer - `(pi)/(4)-x` | |
| 245. |
सरलतम रूप व्यक्त कीजिए - `tan^(-1)((cosx-sinx)/(cosx+sinx)), 0 lt x lt pi`. |
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Answer» यहाँ `tan^(-1)((cosx-sinx)/(cosx+sinx))` `=tan^(-1)((1-tanx)/(1+tanx))` `=tan^(-1){tan((pi)/(4)-x)}` `=(pi)/(4)-x.` जो कि अभीष्ट सरलतम रूप है। |
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| 246. |
`cot^(-1)(-(1)/(5))` का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `(3)/(5)` | |
| 247. |
`sin[2 cos^(-1)(-(3)/(5))]` का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `-(24)/(25)` | |
| 248. |
सिद्ध कीजिए- `tan^(-1)""(1)/(7)+tan^(-1)""(1)/(13)=tan^(-1)""(2)/(9)` |
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Answer» बायाँ पक्ष `=tan^(-1)""(1)/(7)+tan^(-1)""(1)/(13)` `=tan^(-1)""[((1)/(7)+(1)/(13))/(1-(1)/(7)xx(1)/(13))]=tan^(-1)[((13+7)/(91))/((91-1)/(91))]` `tan^(-1)""(2)/(9)=` दायाँ पक्ष |
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| 249. |
माना `tan^(-1)(sqrt(1+x^(2))+x)=(pi)/(4)+(1)/(2)tan^(-1)x.` |
| Answer» Correct Answer - माना `x=tan theta` | |
| 250. |
यदि `cos^(-1)""(x)/(a)+cos^(-1)""(y)/(b)=theta"` तो` "(x^(2))/(a^2)+(y^(2))/(b^(2))` बराबर है-A. `(xy)/(ab)costheta+cos^(2)theta`B. `(2xy)/(ab)costheta+cos^(2)theta`C. `(2xy)/(ab)costheta+sin^(2)theta`D. `(xy)/(ab)costheta+sin^(2)theta` |
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Answer» Correct Answer - C |
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