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InterviewSolution
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. |
`"cosec"theta+cottheta=sqrt(3)` |
| Answer» `theta=2npi+pi/3, theta=(2n-1)pi` | |
| 2. |
`3(sintheta-costheta)^(4)+6(sintheta+costheta)^(2)+4(sin^(6)theta+cos^(6)theta)=?` |
| Answer» Correct Answer - d | |
| 3. |
`sqrt(3)costheta+sintheta=2` |
| Answer» `theta=2npi+pi/6` | |
| 4. |
Prove that : `(sintheta)/(1+costheta)+(1+costheta)/(sintheta)=2"cosec"theta` |
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Answer» LHS `=(sintheta)/(1+ costheta) + (1+costheta)/(sintheta)= (sin^(2)theta + (1+costheta)^(2))/((1+costheta).sintheta)` `=(sin^(2)theta+1+cos^(2)theta+2 costheta)/((1+costheta). Sintheta)` `=(sin^(2)theta+ cos^(2)theta)+1+2costheta/((1+costheta).sintheta)` `=(1+1 + 2 costheta)/((1+costheta).sintheta)` `=(2(1+costheta))/((1+costheta)sintheta)= 2/sintheta` `=2"cosec"theta`= R.H.S Hence proved. |
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| 5. |
`sintheta=-1/2` and `costheta=-sqrt(3)/2` |
| Answer» `theta=2npi+(5pi)/(12), theta=2npi-(13pi)/(12)` | |
| 6. |
The number of solutions of the equation `sintheta+costheta=2` are:A. 1B. 2C. 0D. infinite |
| Answer» Correct Answer - c | |
| 7. |
Find the general values of `theta` from the following equations: i) `sin3theta=cos3theta` ii) `2cos^(2)theta=1-2sinthetacostheta` iii) `"cosec"^(2)theta+2"cosec"theta-3=0` |
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Answer» i) `theta=1/3(npi+pi/4)`, ii) `theta=1/2(npi-pi/4)` iii) `theta=npi+(-1)^(n)pi/2, theta=npi+(-1)^(n), alpha` where `sinalpha=-1/3` |
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| 8. |
If `sintheta-3sin2theta+sin3theta=costheta-3cos2theta+cos3theta`, then one of the general value is:A. `(npi)/(2)+pi/8`B. `npi+pi/8`C. `npi-pi/8`D. `(npi)/(2)-pi/8` |
| Answer» Correct Answer - a | |
| 9. |
Find the general values of `theta` from the following equations: i) `sintheta=-sqrt(3)/(2)` ii) `sectheta=-sqrt(2)` iii) `cottheta=-1/sqrt(3)`, iv) `"cosec"theta=-2` |
| Answer» i) `theta=npi-(-1)^(n)pi/3`, ii) `theta=2npi+-(3pi)/(4)` , iii)` theta=npi-pi/3`, iv) `theta=npi-(-1)^(n).pi/6` | |
| 10. |
If `(2+sqrt(3))costheta=1-sintheta`, then one general value is:A. `2npi+pi/2`B. `npi+pi/2`C. `npi+pi/3`D. None of these |
| Answer» Correct Answer - a | |
| 11. |
Solve: `sinmtheta+sinntheta=0` |
| Answer» `theta=(2rpi)/(m+n), theta=((2r+1)pi)/(m-n), r in I` | |
| 12. |
Find the geneal values of `theta` from the following equations: i) `sintheta=sqrt(3)/2`, ii) `costheta=1/sqrt(2)`, iii) `tantheta=sqrt(3)`, iv) `sectheta=2/sqrt(3)`, v) `cottheta=1/sqrt(3)`, vi) `"cosec"theta=sqrt(2)` |
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Answer» (I) `theta=npi+(-1)^(n).pi/3`, ii) `theta=2npi+-pi/4` iii) `theta=npi+pi/3`, iv) `theta=2npi+-pi/6` v) `theta=npi+pi/3`, vi) `theta=npi+(-1)^(n)pi/4`. |
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| 13. |
Solve: `tanptheta+cotqtheta=0` |
| Answer» `theta=(2n+1)/(2(p-q))pi` | |
| 14. |
Evaluate the following: i) `sin pi/12` ii) `sin pi/8`, iii) `cos pi/8`, iv) `cos pi/(24` |
| Answer» i) `(sqrt(3)-1)/(2sqrt(2)`, ii) `sqrt(2-sqrt(2))/(2)`, iii) `sqrt(2+sqrt(2))` , iv) `sqrt(4+sqrt(6)+sqrt(2))/(8)` | |
| 15. |
In `DeltaABC`, a=3,b=4,c=2, then prove that: `cosA/2 = (3sqrt(6))/(8)` |
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Answer» `s=(a+b+c)/(2)=(3+4+2)/(2)=9/2` and `cosA/2 = sqrt(s(s-a))/(bc)` `=sqrt((9/2(9/2-3))/(4 xx2))` `=sqrt((9 xx 3))/sqrt(4 xx 4 xx 2)=(3sqrt(6))/(8)` Hence Proved. |
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| 16. |
Find the general value of `theta` from the equaion `sqrt(3)sintheta+costheta=1` |
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Answer» `sqrt(3)sintheta+costheta=1` Divide both sides by `sqrt((sqrt(3))^(2)+1^(2))=2` `sqrt(3)/2sintheta+1/2costheta=1/2` `rArr sinpi/3sintheta+cospi/3= cospi/3` `rArr cos(theta-pi/3)=cospi/3` `rArr theta-pi/3=2npi+-pi/3` `rArr theta=2npi+-pi/3+pi/3` `rArr theta=2npi+(2pi)/(3)` or `theta=2npi`. Ans. |
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| 17. |
Find the general values of `theta` which satisfy the following equations: i) `sintheta=1/sqrt(2)` , ii) `costheta=1/2` , (iii) `tantheta=sqrt(3)`, iv) `cottheta=1` , v) `sectheta=2/sqrt(3)`, vi) `"cosec"theta=sqrt(2)` |
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Answer» i) `sintheta=1/sqrt(2)=sinpi/4` `therefore theta=npi+(-n)^(n).pi/4`. Ans. ii) `costheta=1/2=cospi/3` `therefore theta=2npi+-pi/3`. Ans. iii) `tantheta=sqrt(3)=tanpi/3` `rArr theta=npi+pi/3` iv) `cottheta=1` `rArr tantheta=1=tanpi/4` `therefore theta=npi+pi/4`. v) `sectheta=2/sqrt(3)` `rArr costheta=sqrt(3)/2=cospi/6` vi) `"cosec"theta=sqrt(2)` `rArr sintheta=1/sqrt(2)=sinpi/4` `rArr theta=npi+(-1)^(n)pi/4`. Ans. |
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| 18. |
If `"cosec"A=-sqrt(2)` and `(3pi)/(2) lt A lt 2pi`, find the value of `(tanA+"cosec"A+1)/(cotA-"cosecA"+1)`. |
| Answer» Correct Answer - `-1` | |
| 19. |
If `tanA = 12/5` and `pi lt A lt (3pi)/(2)`, find the values of the following: i) `sin2A` ii) `cos2A` iii) `tan2A` |
| Answer» i) `120/169`, ii) `-119/169`, iii) `-120/119` | |
| 20. |
In `DeltaABC`, `a:b:c=15:7:13`, find cosA. |
| Answer» Correct Answer - `-1/26` | |
| 21. |
A triangle side are few `7 cm, 4sqrt3 cm` and` sqrt(13) cm` then the smallest angle is |
| Answer» `angleC=30^(@)` | |
| 22. |
The solution set of the equation `4 sintheta.costheta-2costheta-2sqrt(3)sintheta +sqrt(3)=0` in the interval (0, 2T)ist (B) 13, 3 j (A) 4 4 (C) 4 3, 3 (D) All solutions of the equation 2 sino tann 0 ere obtained by taking a integral valuessof m and n inA. `(3pi)/(4),(7pi)/(4)`B. `(pi/3), (5pi)/(3)`C. `(3pi)/(4), (7pi)/(4), (pi)/(3), (5pi)/(3)`D. None of these |
| Answer» Correct Answer - d | |
| 23. |
Prove that: `sqrt((1-sinA)/(1+sinA))={(secA-tanA,if -pi/2 lt A lt pi/2),(tanA-secA ,if pi/2 lt A lt (3pi)/2):}` |
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Answer» LHS`=sqrt((1-sinA)/(1+sinA))=sqrt((1-sinA,1-sinA)/(1+sinA)), (1-sinA)(1-sinA)` `=sqrt((1-sinA)^(2)/(1-sin^(2)A))= sqrt((1-sinA)^(2)/(cos^(2)A))=(1-sinA)/(|cosA|)` `{{:((1-sinA)/(cosA)"," if -pi/2 lt "A" lt pi/2),((1-sinA)/(-cosA)"," if pi/2 lt Alt (3pi)/2):}` `{{:(1/(cosA)-(sinA)/(cosA),if-pi/2ltAltpi/2),(-1/(cosA)+(sinA)/(cosA),if pi/2ltAlt(3pi)/2):}` `{{:(secA-tanA,if-pi/2ltAltpi/2),(-secA+tanA,ifpi/2ltAlt(3pi)/2):}` =RHS |
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| 24. |
If `tanA=2, secB=-5/3`, where `pi lt A lt (3pi)/(2)`, `pi/2 lt B lt pi`, then i) Find the value of `tan(A+B)`, ii) find the quadrant in which (A+B) terminates. |
| Answer» i) `2/11` ii) First quadrant | |
| 25. |
Prove that `tan 56^@ = (cos11^@ + sin11^@)/(cos11^@ - sin11^@) ` |
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Answer» LHS `=(cos11^(@)+sin11^(@))/(cos11^(@)-sin11^(@))` `=(1+tan11^(@))/(1-tan11^(@))`(Divide `N^(r )` and `D^(r ) "by" cos x)` `=(tan45^(@)+tan11^(@))/(1-tan45^(@)tan11^(@))` `=tan(45^(@)+11^(@))=tan56^(@)` `tan(90-34^(@))=cot34^(@)=`RHS Hence Proved. |
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| 26. |
Prove that: `t a n 70^0=t a n 20^0+2t a n 50^0dot` |
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Answer» `70^(@)=20^(@)+50^(@)` `rArr tan70^(@)=tan(20^(@)+50^(@))` `rArr tan70^(@)-tan70^(@)tan20^(@)tan50^(@)=tan20^(@)+tan50^(@)=tan20^(@)+tan50^(@)` `tan70^(@)-cot20^(@).1/(cot20^(@))tan50^(@)=tan20^(@)+tan50^(@)` `rArr tan70^(@)-tan50^(@)=tan20^(2)+tan50^(@)` `rArr tan70^(@)=tan20^(@)+2tan50^(2)` Hence proved. |
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| 27. |
Prove that: `tan(pi/4+A). Tan((3pi)/4)+A=-1` |
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Answer» LHS `=tan(pi/4+A).tan((3pi)/4+A)` `=tan(pi/4+A).tan{pi/2+(pi/4+A)}` `tan(pi/4+A).{-cot(pi/4+A)}` `=-tan(pi/4+A).1/(tan(pi/4+A))` `=-1`=RHS Hence proved. |
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| 28. |
`cosA+cos(12 0^(@)+A)+cos(12 0^(@)-A)=` |
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Answer» LHS`=cosA+cos(120^(@)+A)+cos(120^(@)-A)` `=cosA+(cos120^(@)cosA-sin120^(@)sinA) + cos120^(@)cosA+sin120^(@)sinA` `=cosA+2cos120^(@)cosA` `=cosA+2cos(180^(@)-60^(@))cosA` `cosA-2(1/2)cosA` `=cosA-cosA=0=` RHS Hence proved |
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| 29. |
Prove that: `cos175^(@)+cos65^(@)+cos55^(@)=0` |
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Answer» LHS `=cos175^(@)+cos65^(@)+cos55^(@)` `=2cos120^(@)cos55^(@)+cos55^(@)` `=2cos(180^(@)-60^(@))cos55^(@)+cos55^(@)` `=-2cos60^(@)+cos55^(@)` `=-2(1/2)cos55^(@)+cos55^(@)` `-cos55^(@)+cos55^(@)=0` =RHS Hence Proved. |
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| 30. |
Prove that: `sin(A+2B)sinA-sinBsin(2A+B)sinB=sin(A+B)sin(A-B)` |
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Answer» LHS. `=sin(A+2B)sinA-sin(2A+B)sinB` `=1/2[{cos(A+2B-A)-cos(A+2B+A)}-{cos(2A+B-B)-cos(2A+B+B)}]` `=1/2[cos2B-cos(2A+2B)-cos2A+cos(2A+2B)]` `=1/2(cos2B-cos2A)` and RHS. `=sin(A+B)sin(A-B)` `=1/2.2sin(A+B)sin(A-B)` `=1/2[cos{(A+B)-(A-B)}-cos{(A+B)+(A-B)}]` `1/2(cos2B-cos2A)` `therefore` L.H.S=R.H.S Hence Proved. |
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| 31. |
In `DeltaABC`, prove that: `(b-c)/(a)=(tanB/2-tanC/2)/(tanB/2+tanC/2)` |
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Answer» LHS `=(b-c)/a=(ksinB-ksinC)/(ksinA)` `=(sinB-sinC)/(sinA)=(2cos(B+C)/(2).sin(B-C)/(2))/(2sinA/2.cosA/2)` `=(cos(B+C)/(2).sin(B-C)/(2))/(sin(180^(@)-(B+C))/(2).cos(180^(@)-(B+C))/(2))` `(cos(B+C)/(2).sin(B-C)/(2))/(sin{90^(@)-(B+C)/(2)}.cos{90^(@)-(B+C)/(2)}` `=(cos(B+C)/(2).sin(B-C)/(2))/(cos(B+C)/(2).sin(B+C)/(2))=(sin(B/2-C/2)/(sin(B/2+C/2)` `=sinB/2cosC/2-cosB/2sinC/2)/(sinB/2cosC/2+cosB/2sinC/2)` `=(tanB/2-tanC/2)/(tanB/2+tanC/2)` (Divide `N^(r)` and `D^(r)` by `cosB/2cosC/2`) =RHS Hence Proved. |
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| 32. |
If n. `sin(A+2B)=sinA`, then prove that: `tan(A+B)=(1+n)/(1-n).tanB` |
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Answer» `nsin(A+2B)=sinA` `rArr n=(sinA)/(sin(A+2B))` `therefore` RHS `=(1+n)/(1-n).tanB=(1+(sinA)/(sin(A+2B)))/(1-(sinA)/(sin(A+2B))).tanB` `(sin(A+2B)+sinA)/(sin(A+2B)-sinA).tanB` `=(2sinA(A+2B+A)/(2)cos(A+2B-A)/(2))/(2cos(A+2B+A)/(2)sin(A+2B-A)/(2)).tanB` `=(sin(A+B)cosBsinB)/(cos(A+B)sinBcosB)` `=tan(A+B)`=LHS Hence Proved. |
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| 33. |
In `DeltaABC`, a=9,b=8 and c=4, prove that: `cosB-2cosC=-4/3` |
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Answer» `cosB=(c^(2)+a^(2)-b^(2))/(2ca)=(4^(2)+9^(2)-8^(2))/(2 xx 4 xx 9)` `=(16+81-64)/(72)=33/72` and `cosC=(a^(2)+b^(2)-c^(2))/(2ab)` `=(9^(2)+8^(2)-4^(2))/(2 xx 9 xx 8)` `=(81 +64-16)/(144)=129/144` Now `cosB-2cosC=33/72-(2 xx 129)/(144)` `=33/72-129/72=-96/72=-4/3`. Hence Proved. |
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| 34. |
In `DeltaABC`, if `(2cosA)/(a)+(cosB)/(b)+(2cosC)/(2)=a/(bc)+b/(ca)`, then `angleA=?`A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
| Answer» Correct Answer - D | |
| 35. |
Find the value of x: i) `xcot(90^(@)+theta)+tan(90^(@)+theta)sintheta+"cosec"(90^(@)+theta)=0` ii) `"cosec"(90^(@)+theta)+x costhetacot(90^(@)+theta)=sin(90^(@)+theta)` |
| Answer» i) `x=sintheta`, ii) `x=tantheta` | |
| 36. |
If in `DeltaABC, a=4,b=6,c=8` then `2cosA+4cosB+cosC=` |
| Answer» Correct Answer - `17/4` | |
| 37. |
In `DeltaABC`, prove that: `(a-b)/c=(sin(A-B)/(2))/(cosC/2)` |
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Answer» LHS `=(a-b)/c` `=(ksinA-ksinB)/(ksinC)=(sinA-sinB)/(sinC)` `=(2cos(A+B)/2.sin(A-B)/(2))/(2sinC/2.cosC/2)` `=(cos(180^(2)-C)/(2)sin(A-B)/2)/(sinC/2cosC/2)` `(therefore A+B+C=180^(@))` `=(cos(90^(2)-C/2)sin(A-B)/(2))/(sinC/2cosC/2)=(sinC/2.sin(A-B)/(2))/(sinC/2.cosC/2)` `=(sin(A-B)/(2))/(cosC/2) = RHS` Hence Proved. |
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| 38. |
Convert the following angles into radians: a) `30^(@)` b) `135^(@)` c) `90^(@)` |
| Answer» a) `pi/6`, b) `(3i)/(4)`, (c ) `pi/2`, d) `(5pi)/(6)`, e) `(25pi)/72`, f) `(pi)/(32)` | |
| 39. |
`tantheta-tantheta/2=sectheta/2` |
| Answer» `theta=2/3(2npi+pi/2), theta=2(2npi-pi/2)` | |
| 40. |
`2sin^(2)pi/6 +"cosec"(7pi)/(6)cos^(2)pi/3=3/2` |
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Answer» L.H.S. `2sin^(2)pi/2+"cosec"^(2)(7pi)/(6)cos^(2)pi/3` `=2(sinpi/6)^(2)+["cosec"(7pi)/(6)cospi/3]^(2)` `=2(sinpi/6)^(2)+["cosec"(pi+pi/6)cospi/3]^(2)` `=2(sin180^(@))/(6))^(2) + ["cosec"(180^(@)+180^(@)/6)cos180^(@)/3]^(2)` `=2(sin 30^(@))^(2) + [-"cosec"30^(@)cos60^(@)]^(2)` `=2(1/2)^(2) + [-2 xx 1/2]^(2)` `=1/2 +1=3/2`= RHS Hence proved. |
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| 41. |
Solve: `tantheta+sectheta=2costheta` |
| Answer» `theta=npi+(-1)^(n)(-pi/2),theta=npi+(-1)^(n)pi/6` | |
| 42. |
`sqrt(2)sectheta+tantheta=1` |
| Answer» `theta=2npi-pi/4` | |
| 43. |
`tantheta+sectheta=sqrt(3)` |
| Answer» `theta=2npi+pi/6,theta=2npi-pi/2` | |
| 44. |
Prove that: `(sec4A-1)/(sec8A-1)=tan2A. cot8A` |
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Answer» LHS= `(sec4A-1)/(sec8A-1)` `=(1/(cos4A)-1)/(cos8A)=((1-cos4A)/(cos4A))/((1-cos8A)/(cos8A))` `(1-cos4A)/(1-cos8A).(cos8A)/(cos4A) (therefore 1-cos2theta=2sin^(2)theta)` `(2sin^(2)2A)/(2sin^(2)4A).(cos8A)/(cos4A)` `=(2sin^(2)2A.cos8A)/(sin4A.(2sin4Acos4A))` `=(2sin^(2)2A.cos8A)/((2sin2Acos2A)sin8A)` `(sin2A)/(cos2A).(cos8A)/(sin8A)` `=tan2A.cot8A` =RHS Hence Proved |
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| 45. |
Prove that: `(1+sinA-cosA)/(1+sinA+cosA)=tanA/2` |
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Answer» LHS `=(1+sinA-cosA)/(1+sinA+cosA)=((1-cosA)+sinA)/((1+cosA)+sinA)` `=(2sin^(2)A/2+2sinA/2cosA/2)/(2cos^(2)A/2+2sinA/2cosA/2)` `=(sinA/2(sinA/2cosA/2))/(cosA/2(cosA/2+sinA/2))` `=(sinA/2)/(cosA/2)=tanA/2`= RHS Hence proved. |
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| 46. |
`cot^(2)pi/6+"cosec"(5pi)/(6) + 3tan^(2)pi/6=6` |
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Answer» LHS `=cot^(2)pi/6 + "cosec"(5pi)/(6)+3tan^(2)pi/6` `=cot^(2)pi/6+"cosec"(pi/6)+3(tanpi/6)^(2)` `=(cot30^(@))^(2) + ("cosec"30^(@))+3(tan 30^(@))^(2)` `=(sqrt(3))^(2)+2+3(1/sqrt(3))^(2)=3+2+1=6` RHS Hence Proved. |
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| 47. |
Solve: `tan^(3)theta-3tantheta=0` |
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Answer» `tan^(3)theta-3tantheta=0` `rArr tantheta(tan^(2)theta-3)=0` `rArr tantheta=0` or `tan^(2)theta-3=0` if `tantheta=0`, then `theta=npi`. If `tan^(2)theta-3=0`, then `tan^(2)theta=3` `=(sqrt(3))^(2)=tan^(2)pi/3` `rArr theta=npi+-pi/3`. Ans. |
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| 48. |
Solve: `sectheta-1=(sqrt(2)-1)tantheta` |
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Answer» `sectheta-1=(sqrt(2)-1)tantheta` `rArr (sectheta-1)^(2)=(sqrt(2)-1)^(2).tan^(2)theta` `=(2+1-2sqrt(2))(sec^(2)theta-1)` `=(3-2sqrt(2))(sec^(2)theta-1)` `rArr (sectheta-1)^(2)=3-2sqrt(2))(sectheta-1)(sectheta+1)` `rArr (3-2sqrt(2)(sectheta-1)(sectheta+1)-(sectheta-1)^(2)=0` `rArr (sectheta-1)[(3-2sqrt(2))(sectheta+1)-(sectheta-1)]=0` `rArr (sectheta-1)[(2-2sqrt(2))sectheta+(4-2sqrt(2))]=0` `rArr (sectheta-1)[(2-2sqrt(2))sectheta-sqrt(2)(-2sqrt(2)+2)]=0` `rArr (sectheta-1)(2-sqrt(2))(sectheta-sqrt(2))=0` `rArr sectheta-1=0` or `sectheta-sqrt(2)=0` Now `sectheta-1=0` `rArr sectheta=1` `rArr costheta=1` `rArr costheta=cos0^(@)` `rArr theta=2npi+-0=2npi` Ans. and `sectheta-sqrt(2)=0` `rArr sectheta=sqrt(2)` `rArr costheta=1/sqrt(2)` `rArr costheta=cosi/4` `rArr theta=2npi+-pi/4` But the given equation is not satisfied by `theta=2npi-pi/4`, so `theta=2npi+pi/4`. Ans. |
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| 49. |
Maximum value of `(3sintheta+4costheta)` is: |
| Answer» Correct Answer - b | |
| 50. |
Maximum value of `(3sintheta+4costheta)` is:A. 5B. `-5`C. 1D. None of these |
| Answer» Correct Answer - A | |