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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.
| 1. |
निम्नलिखित के मुख्य मानो को ज्ञात कीजिए : `cot^(-1)(sqrt(3))` |
| Answer» Correct Answer - `(pi)/(6)` | |
| 2. |
यदि `cot^(-1)(sqrt(cos alpha ))- tan^(-1)(sqrt(cos alpha))=x` तब `sin x=`A. `tan^(1)""(alpha)/(2)`B. `tan alpha `C. `cot^(2)""(alpha)/(2)`D. `cot alpha ` |
| Answer» Correct Answer - A | |
| 3. |
निम्नलिखित के मुख्य मानो को ज्ञात कीजिए : ` sin^(-1)""((-1)/(2))` |
| Answer» Correct Answer - `-(pi)/(6)` | |
| 4. |
सरलतम रूप में लिखिए - `cot^(-1)((1)/(sqrt(x^(2)-1))), |x| gt1` |
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Answer» माना `x=sec theta,` तब `theta=sec^(-1)x` `therefore cot^(-1)((1)/(sqrt(x^(2)-1)))=cot^(-1)((1)/(sqrt(sec^(2)theta-1)))` `=cot^(-1)((1)/(tan theta))` `=cot^(-1)(cot theta)` `=theta=sec^(-1)x` चूँकि अब जो कि अभीष्ट सरलतम रूप है। |
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| 5. |
सरलतम रूप व्यक्त कीजिए - `tan^(-1)((cosx)/(1-sinx))` |
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Answer» यहाँ `tan^(-1)((cosx)/(1-sinx))` `=tan^(-1){(cos^(2).(x)/(2)-sin^(2).(x)/(2))/(cos^(2).(x)/(2)+sin^(2).(x)/(2)-sin.(x)/(2)cos.(x)/(2))}` `=tan^(-1){((cos.(x)/(2)+sin.(x)/(2))(cos.(x)/(2)-sin.(x)/(2)))/((cos.(x)/(2)-sin.(x)/(2))^(2))}` `=tan^(-1){((cos.(x)/(2)+sin.(x)/(2)))/((cos.(x)/(2)-sin.(x)/(2)))}` `=tan^(-1){(1+tan.(x)/(2))/(1-tan.(x)/(2))}` `=tan^(-1){tan((pi)/(4)+(x)/(2))}` `=(pi)/(4)+(x)/(2)`जो कि अभीष्ट सरलतम रूप है। |
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| 6. |
`tan^(-1)((3a^(2)x-x^(3))/(a^(3)-3ax^(2))),agt0 , = (a)/(sqrt(3)) lt x lt (a)/(sqrt(3))` को सरलतम रूप में लिखे | |
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Answer» `x = a tan theta` रखें | `because - (a)/(sqrt(3)) lt x lt (a)/(sqrt(3))` तथा `a gt 0` `therefore - (a)/(sqrt(3)) lt a tan theta lt (a)/(sqrt(3)) rArr - (1)/(sqrt(3)) lt tan theta lt (1)/(sqrt(3))` `rArr - (pi)/(3) lt theta lt (pi)/(3)" "...(1)` `because x = a tan theta therefore tan theta = (x)/(a) rArr theta = tan^(-1)""(x)/(a)" "...(2)` अब, `tan^(-1)[(3a^(2)x - x^(3))/(a^(3) - 3ax^(2))] = tan^(-1)[(3a^(2)a tan theta - a^(3)tan^(3)theta)/(a^(3)-3a a^(2)tan^(2)theta)]` ` = tan^(-1)[(a^(2)(3 tan theta-tan^(3)theta))/(a^(3)(1-3tan^(2)theta))] = tan^(-1)(tan 3theta)` `= (3theta) = 3 tan^(-1)""(x)/(a)" "`[(2) से] |
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| 7. |
`tan^(-1)""(1)/(sqrt(x^(2)-1)),|x|gt1` |
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Answer» `tan^(-1)""(1)/(sqrt(x^(2)-1))" "" माना "{:(x=cosectheta),(impliestheta=cosec^(-1)x):}` `=tan^(-1)(1)/(sqrt(cosec^(2)theta-1))=tan^(-1)((1)/(cottheta))` `=tan^(-1)(tantheta)=theta=cosec^(-1)x` |
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| 8. |
`tan^(-1)""((sqrt(1-cosx))/(1+cosx)),0ltxltpi` |
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Answer» `tan^(-1)(sqrt((1-cosx)/(1+cosx)))` `=tan^(-1)(sqrt((2sin^(2)""(x)/(2))/(2cos^(2)""(x)/(2))))=tan^(-1)((sin""(x)/(2))/(cos""(x)/(2)))` `=tan^(-1)(tan""(x)/(2))=(x)/(2)` |
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| 9. |
`tan^(-1)""(x)/(sqrt(a^(2)-x^(2))),|x|lta` को सरलतम रूप में लिखे | |
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Answer» `x = a sin theta,` रखने पर, `tan^(-1)""(x)/(sqrt(a^(2)-x^(2)))=tan^(-1)""(asintheta)/(sqrt(a^(2) - a^(2) sin^(2)theta)) = tan^(-1) ""(a sin theta)/(sqrt(a^(2) (1-sin^(2)theta)))` ` = tan^(-1)((a sin theta)/(a cos theta)) = tan^(-1) (tan theta) = theta = sin^(-1)""(x)/(a)` |
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| 10. |
`tan^(-1)((cosx - sinx)/(cosx + sinx)), - (pi)/(4)ltxlt(3pi)/(4)` को सरलतम रूप में लिखे | |
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Answer» `tan^(-1)((cosx -sinx)/(cosx + sinx)) = tan^(-1)((1-tanx)/(1+tanx))` ` = tan^(-1)((tan""(pi)/(4)-tanx)/(1+tan""(pi)/(4)*tanx)) = tan^(-1)tan((pi)/(4)-x) = (pi)/(4)-x` |
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| 11. |
`tan^(-1)""(x)/(sqrt(a^(2)-x^(2))),|x|lta` |
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Answer» `tan^(-1)""(x)/(sqrt(a^(2)-x^(2)))=tan^(-1)""(asintheta)/(sqrt(a^(2)-a^(2)sin^(2)theta))` `=tan^(-1)""(asintheta)/(sqrt(a^(2)(1-sin^(2)theta)))" "" माना " {:(x=asintheta),(impliessintheta=(x)/(a)):}` `=tan^(-1)""(asintheta)/(sqrt(a^(2)cos^(2)theta))" "impliestheta=sin^(-1)((x)/(a))` `=tan^(-1)""(asintheta)/(acostheta)=tan^(-1)(tantheta)` `=theta=sin^(-1)""(x)/(a)` |
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| 12. |
`tan^(-1)[2cos(2sin^(-1)""(1)/(2))]` |
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Answer» `tan^(-1)[2cos(2sin^(-1)""(1)/(2))]` `=tan^(-1)[2cos(2sin^(-1)sin""(pi)/(6))]` `=tan^(-1)[2cos(2.(pi)/(6))]` `=tan^(-1)[2.cos""(pi)/(3)]` `=tan^(-1)(2.(1)/(2))=tan^(-1)(1)` `tan^(-1)(tan""(pi)/(4))=(pi)/(4)` |
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| 13. |
`tan^(-1)((cosx-sinx)/(cosx+sinx)),(-pi)/(4)ltxlt(3pi)/(3)` |
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Answer» `tan^(-1)((cosx-sinx)/(cosx+sinx))` `=tan^(-1)(((cosx)/(cosx)-(sinx)/(cosx))/((cosx)/(cosx)+(sinx)/(cosx)))` `=tan^(-1)((1-tanx)/(1+tanx))` `=tan^(-1){tan((pi)/(4)-x)}=(pi)/(4)-x` |
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| 14. |
सरलतम रूप में लिखिए - `sin^(-1)((sinx+cosx)/(sqrt2)),-(pi)/(4) lt x lt(pi)/(4).` |
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Answer» `sin^(-1)((sinx+cosx)/(sqrt2))` `=sin^(-1)[(1)/(sqrt2)sinx+(1)/(sqrt2)cosx]` `=sin^(-1)[sinx cos.(pi)/(4)+cosx sin .(pi)/(4)]` `=sin^(-1)(sin(x+(pi)/(4))),` `[because -(pi)/(4) lt x lt (pi)/(4) rArr 0 lt x +(pi)/(4) lt (pi)/(2)]` `=x+(pi)/(4).` |
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| 15. |
`sin[(1)/(2)cos^(-1)""((4)/(5))]` का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `(1)/(sqrt(10))` | |
| 16. |
`tan""(1)/(2)(cos^(-1)""(sqrt(5))/(3))` का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `(3-sqrt(5))/(2)` | |
| 17. |
सरलतम रूप में लिखिए - `cos^(-1)((sinx+cosx)/(sqrt2)). |
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Answer» `cos^(-1)((sinx+cosx)/(sqrt2))` `=cos^(-1)((1)/(sqrt2)sinx+(1)/(sqrt2)x)` `=cos^(-1)(sin x sin.(pi)/(4)+cosx cos.(pi)/(4))` `=cos^(-1)(cosx cos.(pi)/(4)+sinx sin.(pi)/(4))` `=cos^(-1)(cos(x-(pi)/(4))),` `" "[because (pi)/(4)lt x lt (5pi)/(4) rArr 0 lt x lt -(pi)/(4) lt pi]` `=x-(pi)/(4).` |
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| 18. |
यदि `sin^(-1)x=(1)/(3)" तो " sin^(-1)(2xsqrt(1-x^(2)))` का मान ज्ञात कीजिए। |
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Answer» Correct Answer - `(2)/(3)` |
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| 19. |
`cos^(-1)(cos680^(@))` का मुख्य मान ज्ञात कीजिए । |
| Answer» Correct Answer - `40^(@)` | |
| 20. |
यदि `sin^(-1)""((1)/(3))+cos^(-1)""(1)/(x)=(pi)/(2)`, तब x का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `(1)/(3)` | |
| 21. |
`tan^(-1)""(1)/(2)+tan^(-1)""(1)/(3)` |
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Answer» `tan^(-1)""(1)/(2)+tan^(-1)""(1)/(3)=tan^(-1)""((1)/(2)+(1)/(3))/(1-(1)/(2)xx(1)/(3))` `=tan^(-1)1=tan^(-1)tan""(pi)/(4)=(pi)/(4)` |
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| 22. |
मान ज्ञात कीजिए - `tan^(-1)sqrt(3)` |
| Answer» Correct Answer - `(pi)/(3)` | |
| 23. |
`cos(2cos^(-1)0.8)` का मान है -A. `0.28`B. `0.48`C. `0.60`D. इनमे से कोई नहीं |
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Answer» Correct Answer - A |
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| 24. |
सिद्ध कीजिए कि `tan^(-1)""[(3a^(2)x-x^(3))/(a(a^(2)-3x^(2)))]=3tan^(-1)""(x)/(a)` |
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Answer» माना `tan^(-1)""(x)/(a)=thetaimplies tan theta=(x)/(a)` `:.` दायाँ पक्ष `=3 tan^(-1)""(x)/(a)=3theta` अब, बायाँ पक्ष `tan^(-1)""[(3a^(2)x-x^(3))/(a(a^(2)-3x^(2)))]` =`tan^(-1)""[(3a^(2)x-x^(3))/(a(a^(2)-3x^(2)))]=tan^(-1)[(3(x)/(a)-(x^(3))/(a^(3)))/(1-3(x^(2))/(a^(2)))]` `=tan^(-1)[(3 tan theta -tan^(3)theta)/(1-3tan^(2)theta)]` `=tan^(-1)[tan 3 theta ]=3 theta =3 tan^(-1)""(x)/(a)` `:. tan^(-1)""[(3a^(2)x-x^(3))/(a(a^(2)-3x^(2)))]=3 tan^(-1)""(x)/(a)` |
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| 25. |
x के लिए हल करें : `cot^(-1) 2x + tan^(-1) 3x = (pi)/(4)` |
| Answer» Correct Answer - 3 | |
| 26. |
`sin[cot^(-1){cos(tan^(-1)x)}]` का सरलतम रूप है -A. `sqrt((x^(2)+2)/(x^(2)+1))`B. `(x)/(sqrt(x^(2))+1)`C. `(x)/(sqrt(x^(2)+2))`D. `sqrt((x^(2)+1)/(x^(2)+2))` |
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Answer» Correct Answer - D |
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| 27. |
`cot^(-1)"" (1)/(sqrt(x^(2)-1)),|x| gt 1` का सरलतम रूप में लिखें |
| Answer» Correct Answer - `sec^(-1)x` | |
| 28. |
`tan ^(-1)""(sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x)), (-1)/(sqrt(2)) lexle 1` का सरलतम रूप में लिखें |
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Answer» Correct Answer - `(pi)/(4)-(1)/(2) cos^(-1)x` `(pi)/(4)-(1)/(2) cos^(-1)x` |
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| 29. |
`tan^(-1)((a cos x - b sin x)/(b cos x + a sin x)), (a)/(b) sin x gt-1` को सरलतम रूप में लिखें | |
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Answer» दिया गया व्यंजक `=tan^(-1)[(a cos x-b sinx)/(b cos x + a sinx)] = tan^(-1)[((acosx-bsinx)/(bcosx))/((bcosx + a sinx)/(b cosx))] = tan^(-1)[((a)/(b)-tanx)/(1+(a)/(b)tanx)]` ` = tan^(-1)""(a)/(b)(tanx) = tan^(-1) ""(a)/(b) -x` |
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| 30. |
यदि `sin(sin^(-1)""(1)/(5)+cos^(-1) x)=1 ` तो x का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `(1)/(5)` | |
| 31. |
`sin((pi)/(3)-sin^(-1)(-(1)/(2))) ` का मान है ।A. `(1)/(2)` हैB. `(1)/(3)` हैC. `(1)/(4)` हैD. 1 |
| Answer» Correct Answer - D | |
| 32. |
सिद्ध कीजिए कि ` tan^(-1)""(1)/(2)+tan^(-1)""(1)/(5)+tan^(-1)""(1)/(8)=(pi)/(4)` |
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Answer» `tan^(-1)x +tan^(-1)y+tan^(-1)z=tan^(-1)""(x+y+x-xyz)/(1-xy-yz-zx)` `tan^(-1)""((1)/(2)+(1)/(5)+(1)/(8)-(1)/(2)xx(1)/(5)xx(1)/(8))/(1-(1)/(2)xx(1)/(5)-(1)/(5)xx(1)/(8)-(1)/(8)xx(1)/(2))` ` =tan^(-1)""((40+16+10-1)/(80))/((80-8-2-5)/(80))` ` =tan^(-1)""((65)/(65))=(pi)/(4)` |
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| 33. |
सिद्ध कीजिए कि ` tan^(-1)((x)/(sqrt(a^(2)-x^(2))))=sin^(-1)""(x)/(a)` |
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Answer» हमें सिद्ध करना है कि ` tan^(-1)((x)/(sqrt(a^(2)-x^(2))))=sin^(-1)""(x)/(a)` माना `sin^(0-1)""(x)/(a)=theta` तब `(x)/(a)=sin thetaimpliesx=a sin theta` ` :. (x)/(sqrt(a^(2)-x^(2)))=(a sin theta)/(sqrt(a^(2)-x^(2)sin^(2)theta))` `= (a sin theta)/(a cos theta)=tan theta ` `implies tan^(-1)((x)/(sqrt(a^(2)-x^(2))))= tan^(-1)(tan theta)=theta` तथा `theta=sin^(-1)""(x)/(a)` `implies tan^(-1)((x)/(sqrt(a^(2)-x^(2))))=sin^(-1)""(x)/(a)` |
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| 34. |
यदि `tan^(-1)""(x-1)/(x-2) +tan^(-1)""(x+1)/(x+2)=(pi)/(4) `, तो x का मान ज्ञात कीजिए । |
| Answer» Correct Answer - `pm(1)/(sqrt(2))` | |
| 35. |
सिद्ध कीजिए कि ` cos[tan^(-1){sin(cot^(-1)x)}]=sqrt((x^(2)+1)/(x^(2)+2))` |
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Answer» सर्वप्रथम माना `cot^(-1)x=theta` `implies x= cot theta ` `:. " cosec" theta=sqrt(1+cot^(2)theta)=sqrt(1+x^(2))` ` implies sin theta = (1)/(sqrt(1+x^(2)))` `implies sin (cot^(-1)x)=(1)/(sqrt(1+x^(2)))` `= tan^(-1)[sin(cot^(-1)x)]=tan^(-1)""(1)/(sqrt(1+x^(2)))= phi` (say) तब `, cos[tan^(-1){sin(cot^(-1)x)}]=cos phi" "`......(1) अब , चूँकि `tan^(-1)""(1)/(sqrt(1+x^(2)))=phi` , तब ` tan phi=(1)/(sqrt(1+x^(2)))` `:. sec phi=sqrt(1+tan^(2)phi)=sqrt(1+(1)/((1+x^(2))))=sqrt((x^(2)+1)/(x^(2)+2))` `:. cos phi= sqrt((1+x^(2))/(2+x^(2)))" " `.....(2) अतः समीकरण ( 1 ) व ( 2 ) से , ` cos[tan^(-1){sin(cot^(-1)x)}]=sqrt(x^(2)+1)/(x^(2)+2)` |
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| 36. |
मान ज्ञात कीजिए - `cos(sec^(-1)x+"cosec"^(-1)x),|x| ge 1.` |
| Answer» `cos(sec^(-1)x+"cosec"^(-1)x)=cos((pi)/(2))=0.` | |
| 37. |
मान ज्ञात कीजिए - `sin[sin^(-1)x+cos^(-1)x]` |
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Answer» `sin[sin^(-1)x+cos^(-1)x]=sin.(pi)/(2),` `" "[because sin^(-1)x+cos^(-1)x=(pi)/(2)]` = 1 |
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| 38. |
मान ज्ञात कीजिए - `sin[sin^(-1)x+cos^(-1)x]` |
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Answer» `sin[sin^(-1)x+cos^(-1)x]=sin.(pi)/(2),` `" "[because sin^(-1)x+cos^(-1)x=(pi)/(2)]` = 1 |
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| 39. |
मान ज्ञात कीजिए - `cos[tan^(-1)a+cot^(-1)a]` |
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Answer» `cot[tan^(-1)a+cot^(-1)a]` `=cot.(pi)/(2)=0," "[because tan^(-1)x+cot^(-1)x=(pi)/(2)]` |
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| 40. |
`cos(sin^(-1).(3)/(5)+sin^(-1).(5)/(13))` का मान है - |
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Answer» माना `sin^(-1).(3)/(5)=A` और `sin^(-1).(5)/(13)=B,` जहाँ `A,B in [-(pi)/(2),(pi)/(2)]` तब `sinA=(3)/(5)` और `sinB=(5)/(13).` चूँकि `A, B in [-(pi)/(2),(pi)/(2)]` इसलिए `cosA, cosB gt 0` `therefore" "cosA=sqrt(1-sin^(2)A)` `=sqrt(1-(9)/(25))=(4)/(5)` `cosB=sqrt(1-sin^(2)B)` `=sqrt(1-(25)/(169))=(12)/(13).` अब, `cos(sin^(-1).(3)/(5)+sin^(-1).(5)/(13))=cos(A+B)` `=cosAcos B-sinAsinB` `=((4)/(5)xx(12)/(13))-((3)/(5)xx(5)/(13))` `=(48)/(65)-(15)/(65)` `=(33)/(65)`. |
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| 41. |
`cos[sin^(-1)""(3)/(5)+sin^(-1)""(5)/(13)]` का मान ज्ञात कीजिए । |
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Answer» माना `sin^(-1)""(3)/(5)=A` तथा `sin^(-1)""(5)/(13)=B` ` implies sinA=(3)/(5),sinB=(5)/(13)` `implies cos A=(4)/(5), cosB=(12)/(13)` अब `cos[A+B]=cosA cos B- sinA sinB` `=(4)/(5)xx(12)/(13)-(3)/(5)xx(5)/(13)=(48)/(65)-(15)/(65)=(33)/(65)` |
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| 42. |
सिद्ध कीजिए कि `tan""((1)/(2)cos^(-1)""(sqrt(5))/(3))=(3-sqrt(5))/(2)` |
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Answer» मान ` cos^(-1)""(sqrt(5))/(3)=theta implies cos theta =(sqrt(5))/(3)` `implies tan""(1)/(2)(cos^(-1)""(sqrt(5))/(3))=tan""(theta)/(2)` `=sqrt((1-cos theta )/(1+cos theta))=sqrt((1-sqrt(5)//3)/(1+sqrt(5)//3))=sqrt((3-sqrt(5))/(3+sqrt(5)))` `implies tan""(1)/(2)(cos^(-1)""(sqrt(5))/(3))=(3-sqrt(5))/(2)` |
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| 43. |
सिद्ध कीजिए कि ` sin((1)/(2)cos^(-1)""(4)/(5))=(1)/(sqrt(10))` |
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Answer» मान `cos^(-1)""(4)/(5)=theta implies cos theta =(4)/(5)` ` :. sin((1)/(2)cos^(-1)""(4)/(5))=sin""(theta)/(2)=sqrt((1-cos theta)/(2))=(1)/(10)` `implies sin((1)/(2)cos^(-1)""(4)/(5))=(1)/(sqrt(10))` |
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| 44. |
सिद्ध कीजिए कि - ` sin(cot^(-1)x)=(1)/(sqrt(1+x^(2)))` |
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Answer» मान `cot^(-1)x= theta implies cot theta =x` अब ` sin(cot^(-1)x)=sin theta =(1)/("cosec" theta)=(1)/(sqrt(1+cot^(2)theta))` `implies sin(cot^(-1)x)=(1)/(sqrt(1+x^(2)))` |
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| 45. |
सिद्ध कीजिए कि - `sin(cos^(-1)""(3)/(5))=(4)/(5)` |
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Answer» मान `cos^(-1)""(3)/(5)=thetaimplies cos theta =(3)/(5)` ` :. sin (cos^(-1)""(3)/(5))=sin theta =sqrt(1-cos^(2)theta)=sqrt(1-(9)/(25))` `implies sin(cos^(-1)""(3)/(5))=(4)/(5)` |
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| 46. |
`cos[tan^(-1)""(3)/(4)]=` |
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Answer» मान `tan^(-1)""(3)/(4)=thetaimplies tan theta =(3)/(4)` `implies cos[tan^(-1)""(3)/(4)]=cos theta =(1)/(sec theta )` `= (1)/(sqrt(1+tan^(2)theta))=(1)/(sqrt(1+(9)/(16)))` `:. cos[ tan^(-1)""(3)/(4)]=(4)/(5)` |
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| 47. |
`cot^(-1)[cot(-(pi)/(3))]=` |
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Answer» मान `y=cot^(-1)[cot(-(pi)/(3))]` अब ` cot(-(pi)/(3))=cot((2pi)/(3)-pi)=cot(-pi+(2pi)/(3))=cot""(2pi)/(3)` `:. cot^(-1)[cot(-(pi)/(3))]=cot^(-1)[cot""(2pi)/(3)]` ` implies cot^(-1)[cot(-(pi)/(3))]=(2pi)/(3)` |
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| 48. |
सिद्ध कीजिए - `sin^(-1)(sin""(3pi)/(3))=(pi)/(4)` |
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Answer» मान `y=sin^(-1)(sin""(3pi)/(4))` अब ` sin((3pi)/(4))=sin(pi-(pi)/(4))implies sin((3pi)/(4))=sin""(pi)/(4)` `:. sin^(-1)(sin""(3pi)/(4))=sin^(-1)(sin""(pi)/(4))` `implies sin^(-1)(sin""(3pi)/(4))=(pi)/(4)` |
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| 49. |
सिद्ध कीजिए - ` sec^(-1)(sec""(11pi)/(3))=(pi)/(3)` |
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Answer» `sec^(-1)(sec""(11pi)/(3))=y` अब `sec""(11pi)/(3)=sec(4pi-(pi)/(3))implies sec""(11pi)/(3)=sec""(pi)/(3)` ` :. sec^(-1)(sec""(11pi)/(3))=sec^(-1)(sec""(pi)/(3))` ` sec^(-1)(sec""(11pi)/(3))=(pi)/(3)` |
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| 50. |
निम्न का मान ज्ञात कीजिए । ` cos^(-1)(cos""(7pi)/(6))` |
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Answer» मान ` y= cos^(-1)(cos""(7pi)/(6)), 0 le y le pi` ` y=cos^(-1)[cos(2pi-(5pi)/(6))]` `implies y = cos^(-1)[cos""((5pi)/(6))]` `:. cos^(-1)(cos""(7pi)/(6))=(5pi)/(6)` |
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