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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.
| 251. |
यदि `tan^(-1)""(4)/(3)=theta` तो , ` cos theta` का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `(5)/(sqrt(26))` | |
| 252. |
सिद्ध कीजिए- `tan^(-1)""((cos x-sinx)/(cos x+ sinx ))=(pi)/(4)-x` |
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Answer» यहाँ `(cos x-sinx )/(cosx+sinx)=(cosx(1-(sinx)/(cosx)))/(cosx(1+(sinx)/(cosx)))=(1-tanx)/(1+tanx)=(1-tanx)/(1+1*tanx)` `=(tan""(pi)/(4)-tanx)/(1+tan""(pi)/(4)tanx)=tan""((pi)/(4)-x)` `:.` बायाँ पक्ष `= tan^(-1)""((cosx-sinx)/(cosx+sinx))` ` =tan^(-1)[tan""((pi)/(4)-x)]=(pi)/(4)-x=` दायाँ पक्ष |
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| 253. |
सिद्ध कीजिए कि `tan^(-1)((cosx)/(1+sinx))=(pi)/(4)-(x)/(2)` |
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Answer» बायाँ पक्ष `tan^(-1)((cosx)/(1+sinx))=tan^(-1)[(sin((pi)/(2)-x))/(1+cos((pi)/(2)-x))]` `=tan^(-1)[(2sin((pi)/(4)-(x)/(2))cos((pi)/(4)-(x)/(2)))/(2cos^(2)((pi)/(4)-(x)/(2)))]` `=tan^(-1)[(sin((pi)/(4)-(x)/(2)))/(cos((pi)/(4)-(x)/(2)))]` `=tan^(-1)[tan((pi)/(4)-(x)/(2))]=(pi)/(4)-(x)/(2)` `impliestan^(-1)((cosx)/(1+sinx))=(pi)/(4)-(x)/(2)="दायाँ पक्ष "` |
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| 254. |
सिद्ध कीजिए - `tan^(-1)[(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))]=(pi)/(2)-(x)/(2).` |
| Answer» [परिमेयीकरण करके हल कीजिए ] | |
| 255. |
यदि `tan^(-1)x=theta" तो "sin^(-1)""(2x)/(1+x^(2))`का मान ज्ञात कीजिए। |
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Answer» `tan^(-1)x=theta` `impliesx=tantheta` अब `sin^(-1)""(2x)/(1+x^(2))=sin^(-1)""(2tantheta)/(1+tan^(2)theta)` `=sin^(-1)(sin2theta)=2theta` |
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| 256. |
सिद्ध कीजिए - `cot^(-1)[(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sin x)-sqrt(1-sinx))]=(x)/(2), x in (0,(pi)/(4)).` |
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Answer» `L.H.S."=[(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))]` `=cot^(-1)[(sqrt(cos^(2).(x)/(2)+sin^(2).(x)/(2)+2sin.(x)/(2)cos.(x)/(2))+sqrt(cos^(2).(x)/(2)+sin^(2).(x)/(2)-2sin.(x)/(2)cos.(x)/(2)))/(sqrt(cos^(2).(x)/(2)+sin^(2).(x)/(2)+2sin.(x)/(2)cos.(x)/(2))-sqrt(cos^(2).(x)/(2)+sin^(2).(x)/(2)-2sin.(x)/(2)cos.(x)/(2)))],` `[because cos^(2).(x)/(2)+sin^(2).(x)/(2)-1" और "sinx=2sin.(x)/(2)cos.(x)/(2)]` `=cot^(-1)[(sqrt((cos.(x)/(2)+sin.(x)/(2))^(2))+sqrt((cos.(x)/(2)-sin.(x)/(2))))/(sqrt((cos.(x)/(2)+sin.(x)/(2))^(2))-sqrt((cos.(x)/(2)-sin.(x)/(2))^(2)))]` `=cot^(-1)[((cos.(x)/(2)+sin.(x)/(2))+(cos.(x)/(2)-sin.(x)/(2)))/((cos.(x)/(2)-sin.(x)/(2))-(cos.(x)/(2)-sin.(x)/(2)))],` `[because 0lt (x)/(2)lt(pi)/(4)" और "cos.(x)/(2) gt 0, sin.(x)/(2) gt 0]` `=cot^(-1)[(2cos.(x)/(2))/(2sin.(x)/(2))]` `=cot^(-1)(cot.(x)/(2))` `=(x)/(2)` = R.H.S.`" "`यही सिद्ध करना था। |
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| 257. |
सिद्ध कीजिए कि - `(9pi)/(8)-(9)/(4)sin^(-1)((1)/(3))=(9)/(4)sin^(-1)((2sqrt2)/(3)).` |
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Answer» `L.H.S.=(9pi)/(8)-(9)/(4)sin^(-1)((1)/(3))` `=(9)/(4)[(pi)/(2)-sin^(-1)((1)/(3))]` `=(9)/(4)[cos^(-1)((1)/(3))]," "[because sin^(-1)x+cos^(-1)x=(pi)/(2)]` `=(9)/(4)sin^(-1)(sqrt(1-(1)/(9)))," "[because cos^(-1)x=sin^(-1)sqrt(1-x^(2))]` `=(9)/(4)sin^(-1)(sqrt((8)/(9)))` `=(9)/(4)sin^(-1)((2sqrt2)/(3))` = R.H.S.`" "` यही सिद्ध करना था । |
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| 258. |
मान ज्ञात कीजिए - `"cosec"^(-1)(-1)` |
| Answer» Correct Answer - `-(pi)/(2)` | |
| 259. |
यदि `tan^(-1)""(3)/(4)=x" तो "secx` का मान ज्ञात कीजिए। |
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Answer» Correct Answer - `(5)/(4)` |
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| 260. |
`tan[(1)/(2)cos^(-1)""(sqrt(5))/(3)]` का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `(3-sqrt(5))/(2)` | |
| 261. |
निम्नलिखित समीकरण को हल कीजिए - `tan^(-1)((1-x)/(1+x))=(1)/(2)tan^(-1)x, (x gt0).` |
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Answer» यहाँ `tan^(-1)((1-x)/(1+x))=(1)/(2)tan^(-1)x` `x=tan theta` रखने पर `tan^(-1)[(1-tan theta)/(1+tan theta)]=(1)/(2)tan^(-1)(tan theta)` `rArr tan^(-1)[(tan.(pi)/(4)-tan theta)/(1+tan.(pi)/(4)tan theta)]=(1)/(2)theta` `rArr tan^(-1)[tan((pi)/(4)-theta)]=(1)/(2)theta,` `" "[because tan(A-B)=(tan A-tanB)/(1+tanA tanB)]` `rArr" "(pi)/(4)-theta=(1)/(2)theta` `rArr" "(3)/(2)theta=(pi)/(4)` `rArr" "theta=(pi)/(6)` `rArr" "tan^(-1)x=((pi)/(6))` `rArr" "x=(1)/(3).` |
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| 262. |
सिद्ध कीजिए - `tan^(-1)((sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x)))=(pi)/(4)-(1)/(2)cos^(-1)x, -(1)/(sqrt2)le x le1.` |
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Answer» `L.H.S.=tan^(-1)((sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x)))` `x=cos2theta` रखने पर, `=tan^(-1)[(sqrt(1+cos2theta)-sqrt(1-cos 2theta))/(sqrt(1+cos 2theta)+sqrt(1-cos 2theta))]` `=tan^(-1)[(sqrt(1+(2cos^(2)theta-1))-sqrt(1-(1-2sin^(2)theta)))/(sqrt(1+(2cos^(2)theta-1))+sqrt(1-(1-2sin^(2)theta)))],` `" "[because cos 2theta=2cos^(2)theta-1=1-2sin^(2)theta]` `=tan^(-1)[(sqrt(2cos^(2)theta)-sqrt(2sin^(2)theta))/(sqrt(2cos^(2)theta)+sqrt(2sin^(2)theta))]` `=tan^(-1)[(sqrt2cos theta-sqrt2sin theta)/(sqrt2cos theta+sqrt2sin theta)]` `=tan^(-1)[(cos theta=sin theta)/(cos theta+sin theta)]` `=tan^(-1)[(1-tan theta)/(1+tan theta)],` [अंश और हर को `cos theta` से भाग देने पर ] `=tan^(-1)[(tan.(pi)/(4)-tantheta)/(1+tan.(pi)/(4).tan theta)],` `" "[because tan (A-B)=(tanA-tanB)/(1+tan Atan B)]` `=tan^(-1)[tan((pi)/(4)-theta)]` `=(pi)/(4)-theta` `=(pi)/(4)-(1)/(2)cos^(-1)x` `[because x=cos 2theta rArr 2 theta cos^(-1)x rArr theta=(1)/(2)cos^(-1)x]" "=(x)/(2)` = R.H.S. यही सिद्ध करना था । |
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| 263. |
मान ज्ञात कीजिए - `cos^(-1)""(-(1)/(sqrt(2)))` |
| Answer» Correct Answer - `(3pi)/(4)` | |
| 264. |
`tan^(-1)""((sqrt(1+x^(2))-1)/(x))` का मान `tan^(-1) ` के पदों में ज्ञात कीजिए । |
| Answer» Correct Answer - `(1)/(2) tan^(-1)x` | |
| 265. |
मूल्यांकन कीजिए `sin[2cos^(-1)(-(3)/(5))]` |
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Answer» माना `cos^(-1)(-(3)/(5))=theta`, जहाँ `theta=[0, pi]` तब `" "cos theta=-(3)/(5)` चूँकि `" "theta in [0,pi],` तब `sin theta gt 0` `therefore" "sin theta =sqrt(1-cos^(2)theta)` `=sqrt(1-(-(3)/(5))^(2))=sqrt(1-(9)/(25))` `=sqrt((25-9)/(25))=sqrt((16)/(25))=(4)/(5)` |
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| 266. |
यदि `cos^(-1)x=(pi)/(3)" तो "sin^(-1)x` का मान ज्ञात कीजिए। |
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Answer» Correct Answer - `(pi)/(6)` |
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| 267. |
सिद्ध कीजिए - `tan^(-1)sqrtx=(1)/(2)cos^(-1)((1-x)/(1+x)),x in (0,1)`. |
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Answer» माना `" "tan^(-1)sqrtx=theta, theta in (-(pi)/(2),(pi)/(2))` तब `" "sqrtx=tan theta` `rArr" "x=tan^(2)theta` हम जानते हैं कि, `cos2 theta=(1-tan^(2)theta)/(1+tan^(2)theta)` `rArr" "cos 2 theta=(1-x)/(1+x)` `rArr" "2theta=cos^(-1)((1-x)/(1+x))` `rArr" "theta=(1)/(2)cos^(-1)((1-x)/(1+x))` `rArr" "tan^(-1)sqrtx=(1)/(2)cos^(-1)((1-x)/(1+x))` यही सिद्ध करना था। |
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| 268. |
मान ज्ञात कीजिए - ` sec^(-1)(-sqrt(2))` |
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Answer» Correct Answer - `(3pi)/(4)` `sec^(-1)(-sqrt(2))=x` `implies sec x =-sqrt(2)=-sec""(pi)/(4)=sec(pi-(pi)/(4))=sec""(3pi)/(4)` `implies x=(3pi)/(4)` |
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| 269. |
निम्नलिखित के मुख्य मान ज्ञात कीजिए : `sin^(-1)((1)/(sqrt2))` |
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Answer» Correct Answer - (i) `(pi)/(4)` (ii) `(pi)/(6)` (iii) `(pi)/(3)` (iv) `(pi)/(4)` (v) 0 (vi) `(pi)/(4)` |
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| 270. |
मूल्यांकन कीजिए `sin{(pi)/(3)-sin^(-1)(-(1)/(2))}` |
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Answer» माना `sin[(pi)/(3)-sin^(-1)(-(1)/(2))]` `=sin[(pi)/(3)-{-sin^(-1).(1)/(2)}].` `" "[because sin^(-1)(-x)=-sin^(-1)x]` `=sin[(pi)/(3)+sin^(-1)(sin.(pi)/(6))]` `" "[because (pi)/(6) in [-(pi)/(2),(pi)/(2)]]` `=sin[(pi)/(3)+(pi)/(6)]` `=sin.(pi)/(2)=1.` |
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| 271. |
यदि `tan^(-1)""(x-1)/(x-2)+tan^(-1)""(x+1)/(x+2)=(pi)/(4)`, तो x का मान ज्ञात कीजिए। |
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Answer» `tan^(-1)""(x-1)/(x-2)+tan^(-1)""(x+1)/(x+2)=(pi)/(4)` `impliestan^(-1)""((x-1)/(x-2)+(x+1)/(x+2))/(1-(x-1)/(x-2).(x+1)/(x+2))=tan^(-1)1` `implies((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)=1implies2x^(2)-4=-3` `implies2x^(2)=1impliesx^(2)=(1)/(2)impliesx=+-(1)/(sqrt2)` |
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| 272. |
`tan^(-1)sqrtx=(1)/(2)cos^(-1)((1-x)/(1+x)),x""in[0,1]` |
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Answer» माना `sqrt3=tanthetaimpliesx=tan^(2)theta` `R.H.S.=(1)/(2)cos^(-1)((1-x)/(1+x))" "x"in[0,1]` `=(1)/(2)cos^(-1)((1-tan^(2)theta)/(1+tan^(2)theta))` `=(1)/(2)cos^(-1)(cos2theta)=(1)/(2)(2theta)` `=theta=tan^(-1)sqrt(x)=L.H.S.` |
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| 273. |
निम्नलिखित के मुख्य मान ज्ञात कीजिए : `sin^(-1)((-sqrt3)/(2))` |
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Answer» Correct Answer - (i) `-(pi)/(4)` (ii) `(5pi)/(6)` (iii) `(2pi)/(3)` (iv) `(5pi)/(6)` (v) `-(pi)/(4)` (vi) `-(pi)/(2)`. |
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| 274. |
मूल्यांकन कीजिए `sin(cot^(-1)x)` |
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Answer» माना `cot^(-1)x=theta`, जहाँ `theta in (0,pi)` तब `cot theta=x` चूँकि `theta in (0,pi)` इसलिए `sin theta gt 0` `therefore" "sin theta=(1)/("cosec" theta)=(1)/(sqrt(1+cot^(2)theta))` `=(1)/(sqrt(1+x^(2)))` `rArr" "sin(cot^(-1)x)=(1)/(sqrt(1+x^(2)))` |
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| 275. |
`tan(sin^(-1)""(3)/(5)+cot^(-1)""(3)/(2))` |
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Answer» `tan(sin^(-1)""(3)/(5)+cot^(-1)""(3)/(2))` `=tan[tan^(-1)""((3)/(5))/(sqrt(1-((3)/(5))^(2)))+tan^(-1)""(2)/(3)]` `=tan[tan^(-1)""(3)/(4)+tan^(-1)""(2)/(3)]` `tan[tan^(-1)""((3)/(4)+(2)/(3))/(1-(3)/(4).(2)/(3))]=((9+8)/(12))/((12-6)/12)=(17)/(6)` |
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| 276. |
`tan^(-1)""(1)/(5)+tan^(-1)""(1)/(7)+tan^(-1)""(1)/(3)+tan^(-1)""(1)/(8)=(pi)/(4)` |
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Answer» `L.H.S.=tan^(-1)""(1)/(5)+tan^(-1)""(1)/(7)+tan^(-1)""(1)/(3)+tan^(-1)""(1)/(8)` `=tan^(-1)""((1)/(5)+(1)/(7))/(1-(1)/(5)xx(1)/(7))+tan^(-1)""((1)/(3)+(1)/(8))/(1-(1)/(3)xx(1)/(8))` `=tan^(-1)""((7+5)/(35))/((35-1)/(35))+tan^(-1)""((8+3)/(24))/((24-1)/(24))` `=tan^(-1)""(6)/(17)+tan^(-1)""(11)/(23)` `=tan^(-1)""((6)/(17)+(11)/(23))/(1-(6)/(17)xx(11)/23)=tan^(-1)""((138+187)/(391))/((391-66)/(391))` `=tan^(-1)""(325)/(325)=tan^(-1)1=(pi)/(4)=R.H.S.` |
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| 277. |
निम्नलिखित के मुख्य मान ज्ञात कीजिए। (i) `sin^(-1)""(1)/(2)` (ii) `tan^(-1)""(1)/(sqrt3)` (iii) `cot^(-1)(-sqrt3)` |
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Answer» (i) `sin^(-1)""(1)/(2)=sin^(-1)""(sin""(pi)/(6))=(pi)/(6)` (ii) `tan^(-1)""(1)/(sqrt3)=tan^(-1)(tan""(pi)/(6))=(pi)/(6)` (iii) `cot^(-1)(sqrt-3)=pi-cot^(-1)(sqrt3)=pi-cot^(-1)(cot""(pi)/(6))=pi-(pi)/(6)=(5pi)/(6)` |
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| 278. |
निम्नलिखित का मान ज्ञात कीजिए - `sec(sec^(-1).(27)/(10)).` |
| Answer» `sec(sec^(-1).(27)/(10))=(27)/(10)," "[because(27)/(10) gt 1]` | |
| 279. |
`cos^(-1)""(4)/(5)+cos^(-1)""(12)/(13)=cos^(-1)""(33)/(65)` |
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Answer» `L.H.S.cos^(-1)""(4)/(5)+cos^(-1)""(12)/(13)` `=cos^(-1)[(4)/(5)xx(12)/(13)-sqrt(1-((4)/(5))^(2))sqrt(1-((12)/(13))^(2))]` `=cos^(-1)[(48)/(65)-(3)/(5)xx(5)/(13)]=cos^(-1)((48)/(65)-(15)/(65))` `=cos^(-1)""(33)/(65)=R.H.S.` |
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| 280. |
निम्नलिखित व्यंजकों के मान ज्ञात कीजिए - `cos^(-1)(cos.(2pi)/(4))` |
| Answer» `cos^(-1)(cos.(2pi)/(4))=(2pi)/(4)," "[because (2pi)/(4) in [0, pi]]` | |
| 281. |
`sin((pi)/(3)-sin^(-1)(-(1)/(2)))` का मान है :A. `(1)/(2)`B. `(1)/(3)`C. `(1)/(4)`D. 1 |
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Answer» Correct Answer - D `sin[(pi)/(3)-sin^(-1)(-(1)/(2))]` `=sin[(pi)/(3)+sin^(-1)""(1)/(2)]` `=sin[(pi)/(3)+sin^(-1)(sin""(pi)/(6))]` `=sin[(pi)/(3)+(pi)/(6)]` `=sin""(pi)/(2)=1` |
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| 282. |
निम्नलिखित का मान ज्ञात कीजिए - `sin(sin^(-1)((-1)/(2)))` |
| Answer» `sin(sin^(-1)(-(1)/(2)))=-(1)/(2)," "[because-(1)/(2) in [-1,1]]` | |
| 283. |
`tan^(-1)sqrt3-cot^(-1)(-sqrt3)` का मान है :A. `pi`B. `-(pi)/(2)`C. 0D. `2sqrt3` |
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Answer» Correct Answer - B `tan^(-1)sqrt3-cot^(-1)(-sqrt3)` `=tan^(-1)sqrt3-(pi-cot^(-1)sqrt3)` `=tan^(-1)sqrt3-pi+cot^(-1)sqrt3` `=(tan^(-1)sqrt3+cot^(-1)sqrt3)-pi` `=(pi)/(2)-pi=-(pi)/(2)` |
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| 284. |
`tan^(-1)(tan""(7pi)/(6))` |
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Answer» `tan^(-1)(tan""(7pi)/(6))=tan^(-1)[tan(pi+(pi)/(6))]` `=tan^(-1)(tan""(pi)/(6))=(pi)/(6)` |
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| 285. |
निम्नलिखित व्यंजकों के मान ज्ञात कीजिए - `cos^(-1)(cos.(2pi)/(3))+sin^(-1)(sin.(2pi)/(3))` |
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Answer» चूँकि `cos^(-1)` की मुख्य शाखा का परिसर `[0,pi]` और `sin^(-1)` का `[-(pi)/(2),(pi)/(2)]` होता है। `therefore" "cos^(-1)(cos.(2pi)/(3))+sin^(-1)(sin.(2pi)/(3))` `=(2pi)/(3)+sin^(-1)[sin(pi-(pi)/(3))]` `=(2pi)/(3)+sin^(-1)(sin.(pi)/(3))," "[because sin(pi-theta)=sin theta]` `=(2pi)/(3)+(pi)/(3)=(3pi)/(3)=pi.` |
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| 286. |
निम्नलिखित व्यंजकों के मान ज्ञात कीजिए - `tan^(-1)(tan.(3pi)/(4))` |
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Answer» चूँकि `tan^(-1)` की मुख्य शाखा का परिसर `(-(pi)/(2),(pi)/(2))` होता है। `therefore" "tan^(-1)(tan.(3pi)/(4))ne(3pi)/(4)," "[because (3pi)/(4) cancel in (-(pi)/(2),(pi)/(2))]` अब , `tan^(-1)(tan.(3pi)/(4))` `=tan^(-1)[tanpi-((pi)/(4))]," "[because (3pi)/(4)=pi-(pi)/(4)]` `=tan^(-1)[-tan((pi)/(4))],` `" "[because tan(pi-theta)=-tan theta]` `=tan^(-1)[tan(-(pi)/(4))]," "[because tan(-theta)=-tan theta]` `=-(pi)/(4) in (-(pi)/(2),(pi)/(2))," "[because tan^(-1)(tan theta)=theta]` `therefore" "tan^(-1)(tan.(3pi)/(4))=-(pi)/(4).` |
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| 287. |
निम्नलिखित व्यंजकों के मान ज्ञात कीजिए - `cos^(-1)(cos.(13pi)/(6))` |
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Answer» चूँकि `cos^(-1)` की मुख्य शाखा का परिसर `[0,pi]` होता है। `therefore" "cos^(-1)(cos.(13pi)/(6))ne(13pi)/(6)," "[because(13pi)/(6) cancel in [0,pi]]` अब, `cos^(-1)(cos.(13pi)/(6))` `=cos^(-1)[cos(2pi+(pi)/(6))]," "[because(13pi)/(6)=2pi+(pi)/(6)]` `=cos^(-1)(cos.(pi)/(6))," "[because cos(2pi+theta)=cos theta]` `=(pi)/(6)in [0,pi]," "[because cos^(-1)(cos theta)=theta]` `therefore cos^(-1)(cos.(13pi)/(6))=(pi)/(6).` |
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| 288. |
मान निकालें : `cos^(-1)cos((13pi)/(6))` |
| Answer» Correct Answer - `(pi)/(6)` | |
| 289. |
`cos^(-1)(cos""(13pi)/(6))` |
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Answer» `cos^(-1)(cos""(13pi)/(6))=cos^(-1)[cos(2pi+(pi)/(6))]` `=cos^(-1)(cos""(pi)/(6))=(pi)/(6)` |
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| 290. |
`cos^(-1)(cos""(13pi)/(6))` का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `(pi)/(6)` | |
| 291. |
सिद्ध कीजिए कि `tan^(-1)""(1)/(4)+tan^(-1)""(2)/(9)=(1)/(2)cos^(-1)""(3)/(5)` |
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Answer» बायाँ पक्ष `=tan^(-1)""(1)/(4)+tan^(-1)""(2)/(9)` `= tan^(-1)(((1)/(4)+(2)/(9))/(1-(1)/(4) xx (2)/(9)))=tan^(-1)""((17)/(34))=tan^(-1)""(1)/(2)` दायें पक्ष में , माना ` (1)/(2) cos^(-1)""(3)/(5)=thetaimplies cos2 theta=(3)/(5)` `:. tan theta=sqrt((1-cos2theta)/(1+cos 2 theta))=sqrt((1-3//5)/(1+3//5))` `tan theta=(1)/(2)implies theta=tan^(-1)""(1)/(2)` `implies` बायाँ पक्ष = दायाँ पक्ष |
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| 292. |
सिद्ध कीजिए कि `cos^(-1)""((63)/(65))+2 tan^(-1)""(1)/(5))=sin^(-1)""(3)/(5)` |
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Answer» हम जानते है कि ` 2tan^(-1)x=cos^(-1)((1-x^(2))/(1+x^(2)))` ` 2 tan^(-1)""(1)/(5)=cos^(-1)((1-1//25)/(1+1//25))` `= cos^(-1)((24)/(26))=cos^(-1)((12)/(13))` `:.` दिये गये समीकरण से ` cos^(-1)""((63)/(65))+cos^(-1)((12)/(13))` `=cos^(-1)[(63)/(65) xx (12)/(13)-sqrt(1-((63)/(65))^(2))sqrt(1-((12)/(13))^(2))]` `= cos^(-1)[(756)/(845)-(80)/(845)]` ` =cos^(-1)[(676)/(845)]=sin^(-1)sqrt(1-((676)/(845))^(2))` `=sin^(-1)((507)/(845))` `implies cos^(-1)""(63)/(65)+cos^(-1)""(12)/(13)=sin^(-1)""(3)/(5)` |
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