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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. |
Range of `"cosec"^(-1)x` isA. `(-pi/2,pi/2)`B. `[-pi/2,pi/2]`C. `[-pi/2,pi/2]-{0}`D. none of these |
| Answer» Correct Answer - C | |
| 2. |
The value of `"cosec"^(-1)("cosec"(4pi)/3)` isA. `(3pi)/4`B. `pi/4`C. `-pi/4`D. none of these |
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Answer» Correct Answer - B Let `"cosec"^(-1)("cosec"(4pi)/3)=x`, where `x in [-pi/2, pi/2]-{0}`. Then, `"cosec"x="cosec"(4pi)/3="cosec:(pi+pi/3)=-"cosec"pi/3="cosec"(-[i/3)`. `therefore x=-pi/3`. |
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| 3. |
Find the principalvalue of `sec^(-1)((-2)/(sqrt(3)))`A. `pi/6`B. `-pi/6`C. `(5pi)/6`D. `(7pi)/6` |
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Answer» Correct Answer - C Let `sec^(-1)(-2/sqrt(3))=x`, where `x in [0,pi]-{pi/2}` Then, `secx=-2/sqrt(3)=-secpi/6 = sec(pi-pi/6)=sec(5pi)/6 rArr x=(5pi)/6`. |
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| 4. |
The principal value of `"cosec"^(-1)(-sqrt(2))` isA. `-pi/4`B. `(3pi)/4`C. `(5pi)/4`D. none of these |
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Answer» Correct Answer - A Let `"cosec"^(-1)(-sqrt(2))=x`, where `xin [-pi/2,pi/2]-{0}`. Then, `"cosec"x=-sqrt(2)=-"cosec"pi/4="cosec"(-pi/4) rArr x=-pi/4`. |
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| 5. |
Find the principal value of each of the following: i) `cos^(-1)(-1/sqrt(2))`, ii) `cos^(-1)(-1/2)`, iii) `cot^(-1)(-1/sqrt(3))` |
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Answer» i) We know that the range of the principal value of `cos^(-1)` is `[0,pi]`. Let `cos^(-1)(-1/sqrt(2))=theta`. Then, `costheta=-1/sqrt(2)=-cospi/4=cos(pi-pi/4)=cos(3pi)/4`. `therefore theta=(3pi)/4 in [0, pi]`. Hence, the principal value of `cos^(-1)(-1/sqrt(2))` is `(3pi)/4`. ii) We know that the range of the principal value of `cos^(-1)` is `[0,pi]`. Let `cos^(-1)(-1/2)=theta`. Then, `costheta=-1/2=-cospi/3=cos(pi-pi/3)=cos""(2pi)/3`. iii) We know that the range of the principal value of `cot^(-1)` is `(0,pi)`. Let `cot^(-1)(-1/sqrt(3))=theta`. Then, `cottheta=-1/sqrt(3)=-cotpi/3=cot(pi-pi/3)=cot(2pi)/3`. `therefore theta=(2pi)/3 in (0,pi)`. Hence, the principal value of `cot^(-1)(-1/sqrt(3))` is `(2pi)/3.` |
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| 6. |
Find the principal value of each of the following: i) `cot^(-1)(-sqrt(3))` ii) `sec^(-1)(-sqrt(2))` iii) `"cosec"^(-1)(-1)`. |
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Answer» i) We know that the range of the principal value of `cot^(-1)` is `(0,pi)`. Let `cot^(-)(-sqrt(3))=theta`. Then, `cottheta=-sqrt(3)=-cotpi/6=cot(pi-pi/6)=cot(5pi)/6`. `therefore theta=(5pi)/6 in (0, pi)`. Hence, the principal value of `cot^(-1)(-sqrt(3))` is `(5pi)/6`. ii) We know that the range of principal value of `sec^(-1)` is `[0,pi]-{pi/2}`. Let `sec^(-1)(-sqrt(2))=theta`. Then, `sec^(-1)(-sqrt(2))=theta`. Then, `sectheta=-sqrt(2)=-secpi/4 = sec(pi-pi/4)=sec(3pi)/4`. `therefore theta=(3pi)/4 in [0,pi]-{pi/2}`. Hence, the principal value of `sec^(-1)(-sqrt(2))` is `(3pi)/4`. iii) We know that the range of the princiapl value of `"cosec"^(-1)` is `[-pi/2,pi/2]-{0}`. Let `"cosec"^(-1)(-1)=theta`. Then, `"cosec"theta=-1`. `"cosec"theta=-1=-"cosec"pi/2="cosec"(-pi/2)`. `therefore theta=-pi/2 in [-pi/2, pi/2]-{0}`. Hence, the principal value of `"cosec"^(-1)(-1)` is `pi/2`. |
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| 7. |
The principal value of `cot^(-1)(-1)` isA. `-pi/4`B. `pi/4`C. `(5pi)/4`D. `(3pi)/4` |
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Answer» Correct Answer - D Let `cos^(-1)(-1)=x`, where `x in [0,pi]`. Then, `cotx=-1=-cotpi/4=cot(pi-pi/4)=cot(3pi)/4 rArr x=(3pi)/4`. |
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| 8. |
Find the principal valuesof each of the following:`cot^(-1)(-sqrt(3))`(ii) `cot^(-1)(sqrt(3))`A. `-pi/6`B. `pi/6`C. `(7pi)/6`D. `(5pi)/6` |
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Answer» Correct Answer - D Let `cot^(-1)(-sqrt(3))=x`, where `x in [0,pi]` Then, `cotx=-sqrt(3)=-cotpi/6=cot(pi-pi/6)=cot(5pi)/6 rArr x=(5pi)/6`. |
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| 9. |
If `tan^(-1)x=pi/4-tan^(-1)(1/3)` then `x` isA. `1/2`B. `1/4`C. `1/6`D. none of these |
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Answer» Correct Answer - A `pi/4-tan^(-1)1/3=tan^(-1)1=tan^(-1){(1-1/3)/(1+1/3)}=tan^(-1)1/2 rArr x=1/2`. |
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| 10. |
The value of `tan^(-1)(tan(8*pi/6))` isA. `(7pi)/6`B. `(5pi)/3`C. `pi/3`D. none of these |
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Answer» Correct Answer - C Let `tan^(-1)(tan(8pi)/6)=x`, where `x in (-pi/2,pi/2)`. Then, `tanx=tan(8pi)/6 = tan(pi+2pi/6)=tanpi/3 rArr x=pi/3`. |
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| 11. |
Prove that `cos (tan^(-1) (sin (cot^(-1) x))) = sqrt((x^(2) + 1)/(x^(2) + 2))` |
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Answer» Let `cos^(-1)x=theta`. Then, `x=cot(theta)`. `"cosec"theta=sqrt(1+cot^(2)theta)=sqrt(1+x^(2))` `sintheta=1/sqrt(1+x^(2))` `sin(cot^(-1)x)=1/sqrt(1+x^(2))[ therefore theta=cot^(-1)x]` `tan^(-1){sin(cot^(-1)x)}=tan^(-1)1/sqrt(1+x^(2))=phi` (say) `rArr cos[tan^(-1){sin(cot^(-1)x)]=cosphi`………………….(i) Now, `tan^(-1)1/sqrt(1+x^(2))=phi rArr tanphi=1/sqrt(1+x^(2))` `rArr secphi=sqrt(1+tan^(2)phi)= sqrt(1+1/(1+x^(2))=sqrt((2+x^(2))/(1+x^(2))` `rArr cosphi=sqrt(1+x^(2))/(2+x^(2))`...............(ii) From (i) and (ii), we get `cos[tan^(-1){sin(cot^(-1)x)}]=sqrt((x^(2)+1)/(x^(2)+2)]` |
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| 12. |
Prove that: `tan^(-1){pi/4+1/2 cos^(-1)a/b}+tan{pi/4-1/2 cos^(-1)a/b}=(2b)/a` |
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Answer» Let `cos^(-1)a/b)=theta`. Then, `a/b=costheta`. `therefore` LHS `=tan(pi/4+1/1theta}=tan{pi/4-1/2theta}` `=(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/(costheta)=(2b)/a`=RHS Hence, LHS=RHS. |
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| 13. |
Prove that `2tan^(-1)1/x=sin^(-1)((2x)/(x^(2)+1))` |
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Answer» Let `tan^(-1)1/x=theta`. Then, `1/x=tantheta rArr x=cottheta`. `therefore` LHS `=2tan^(-1)1/x=2theta`. RHS `=sin^(-1)(2cottheta)/(cot^(2)theta+1)=sin^(-1)(2tantheta)/(1+tan^(2)theta)` `=sin^(-1)(sin 2theta)=2theta` `therefore` LHS=RHS. Hence, `2tan^(-1)1/x=sin^(-1)((2x)/(x^(2)+1))`. |
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| 14. |
Evaluate: i) `sin{pi/3-sin^(-1)(-1/2)}` ii) `sin(1/2cos^(-1)4/5)` iii) `sin(cot^(-1)x)` iv) `tan1/2(cos^(-1)sqrt(5)/3)` |
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Answer» i) We know that `sin^(-1)(-theta)=-sin^(-1)theta`. `therefore sin{pi/3-sin^(-1)(-1/2)}=sin{pi/3+sin^(-1)1/2}=sin(pi/3+pi/6)[therefore sin^(-1)1/2=pi/6]` `=sin pi/2=1`. ii) Let `cos^(-1)4/5=theta`, where `theta in [0,pi]`. Then, `costheta=4/5`. Since, `theta in [0,pi] rArr 1/2 theta in [0,pi/2] rArr sin1/2theta gt 0`. `therefore sin(1/2cos^(-1)4/5)=sin1/2theta=sqrt((1-costheta)/(2)) = sqrt((1-(4/5))/(2)=1/sqrt(10)`. ii) Let `cot^(-1)x=theta`. Then, `theta in [0,pi]`. `therefore sin(cot^(-1)x)=sin theta gt 0`. Now, `sin theta=1/("cosec"theta)=1/sqrt(1+cot^(2)theta)=1/sqrt(1+x^(2))`. `therefore sin(cot^(-1)x)=sintheta=1/sqrt(1+x^(2))`. iv) Let `cos^(-1)sqrt(5)/3=theta`. Then, `costheta=sqrt(5)/3`, where `theta in [0,pi]`. `therefore tan1/2(cos^(-1)sqrt(5)/3)=tan1/2theta=sqrt((1-costheta)/(1+costheta))=sqrt((1-sqrt(5)//3)/(1+sqrt(5)//3))=sqrt((3-sqrt(5))/(3+sqrt(5)) xx (3-sqrt(5))/(3-sqrt(5)))=(3-sqrt(5))/(2)`. |
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| 15. |
Prove that `tan^(-1)[(sqrt(1+x^2)+sqrt(1-x^2))/(sqrt(1+x^2)-sqrt(1-x^2))]=pi/4+1/2cos^(-1)x^2` |
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Answer» Putting, `x^(2)=cos2theta`, we get `tan^(-1)((sqrt(1+x^(2))+sqrt(1-x^(2)))/(sqrt(1+x^(2))-sqrt(1-x^(2))))=tan^(-1)((sqrt(1+cos2theta)+sqrt(1-cos2theta))/(sqrt(1+cos2theta)-sqrt(1-cos2theta)))` `tan^(-1) (costheta+sintheta)/(costheta-sintheta)=tan^(-1)(1+tantheta)/(1-tantheta)`. [dividing num. and denom. by `cos theta`] `=tan^(-1){tan(pi/4+theta)}=(pi/4+theta)` `=pi/4+1/2cos^(-1)x^(2)` `therefore tan^(-1) ((sqrt(1+x^(2))+sqrt(1-x^(2)))/(sqrt(1+x^(2))-sqrt(1-x^(2))))=pi/4+1/2cos^(-1)x^(2)`. |
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| 16. |
Evaluate: i) `tan{1/2cos^(-1)sqrt(5)/3}`, ii) `tan{2tan^(-1)1/5-pi/4}` |
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Answer» i) Let `cos^(-1)sqrt(5)/3=theta`. Then, `costheta=sqrt(5)/3`. `therefore tan{1/2cos^(-1)sqrt(5)/3}=tantheta/2` `=sqrt((1-costheta)/(1+costheta))=sqrt((1-sqrt(5)/3)/(1+sqrt(5)/3))=sqrt((3-sqrt(5))/(3+sqrt(5))` `=sqrt((3-sqrt(5))/(3+sqrt(5)) xx (3-sqrt(5))/(3+sqrt(5)) = (3-sqrt(5))/sqrt(9-5)` `=(3-sqrt(5))/sqrt(4)=(3-sqrt(5))/(2)` ii) `tan{2tan^(-1)1/5-pi/4}` `=tan{tan^(-1)(2 xx 1/5)/(1-1/25))-tan^(-1)1}` `=tan{tan^(-1)(2/5 xx 25/24)-tan^(-1)1}`. `=tan{tan^(-1)5/12-tan^(-1)1}` `=tan{tan^(-1)(5/12-1)/(1+5/12)}=tan{tan^(-1)(-7/17)}=-7/17`. |
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| 17. |
For the principalvalues, evaluate each of the following :`sin^(-1)(-1/2)+2cos^(-1)(-(sqrt(3))/2)`(ii) `sin^(-1)(-(sqrt(3))/2)+cos^(-1)((sqrt(3))/2)`A. `1/sqrt(26)`B. `5/sqrt(26)`C. `1/sqrt(24)`D. none of these |
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Answer» Correct Answer - C Range of `cos^(-1)` is `[0,pi]`. `cos^(-1)(-sqrt(3)/2)=x rArr cosx=-sqrt(3)/2 =-cospi/6 = cos(pi-pi/6)=cos(5p)/6 rArr x=(5pi)6` `sin^(-1)(-1/2)+2cos^(-1)(-sqrt(3)/2)=-sin^(-1)1/2+2 xx (5pi)/6 = -pi/6+(5pi)/6=(9pi)/6=(3pi)/2`. |
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| 18. |
`tan(1/2cos^(- 1) sqrt(5)/3)`A. `(3-sqrt(5))/(2)`B. `(3+sqrt(5))/(2)`C. `(5-sqrt(3))/(2)`D. `(5+sqrt(3))/(2)` |
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Answer» Correct Answer - A Let `cos^(-1)sqrt(5)/3=theta`. Then, `costheta =sqrt(5)/3`. `tan1/2(cos^(-1)sqrt(5)/3)=tan1/2theta=(sin(theta/2))/(cos(theta/2))`. `=sqrt((1-costheta)/(1+costheta))={(1-sqrt(5)/3)/(1+sqrt(5)/3)}^(1//2)` `((3-sqrt(5))/(3+sqrt(5)))^(1//2)={(3-sqrt(5))/(3+sqrt(5)) xx (3-sqrt(5))/(3-sqrt(5))}^(1//2)=(3-sqrt(5))/(2)`. |
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| 19. |
Prove that: i) `tan^(-1)(sqrt(x)+sqrt(y))/(1-sqrt(xy))=tan^(-1)sqrt(x)+tan^(-1)sqrt(y)` ii) `tan^(-1)(x+sqrt(x))/(1-x^(3//2))=tan^(-1)x+tan^(-1)sqrt(x)` iii) `tan^(-1)(sinx)/(1+cosx)=x/2` |
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Answer» i) Put `sqrt(x)=tantheta` and `sqrt(y)=tanphi` on each side. ii) Put `x=tantheta` and `sqrt(x)=tanphi` on each side. iii) LHS `=tan^(-1){(2sin(x/2)cos(x/2))/(2cos^(2)x/2)}=tan^(-1)(tanx/2)=x/2`= RHS |
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| 20. |
दर्शाइए कि (i) `sin^(-1)(2xsqrt(1-x^(2)))=2sin^(-1),-(1)/(sqrt(2)) le x le (1)/(sqrt(2))` (ii) `sin^(-1)(2xsqrt(1-x^(2)))=2cos^(-1) x,(1)/(sqrt(2)) le x le 1` |
| Answer» Correct Answer - Put `x=sintheta` | |
| 21. |
Find the value of: i) `sin ^(-1)(sin (3pi)/5)`, ii) `cos^(-1)(cos(13pi)/6)`, iii) `tan^(-1)(tan (7pi)/6)`. |
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Answer» i) We know that the principal-value branch of `cos^(-1)` is `[0,pi]`. `therefore cos^(-1)(cos (13pi)/6) ne (13pi)/6`. Now, `cos^(-1)(cos (13pi)/6=cos^(-1){cos(2pi+pi/6)}=cos^(-1){cospi/6}[therefore cos(2pi+theta)=costheta]=pi/6`. Hence, `cos^(-1)(cos (13pi)/6)) =pi/6`. iii) We know that the principal-value branch of `tan^(-1)` is `(-pi/, pi/2)`. `therefore tan^(-1)(tan(7pi)/6)) ne (7pi)/6`. Now, `tan^(-1)(tan(7pi)6) = tan^(-1){tan(pi+pi/6)}` `=tan^(-1)(tanpi/6) [therefore tan(pi+theta)=tantheta]` `=pi/6`. Hence, `tan^(-1)(tan(7pi)/6)=pi/6`. |
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| 22. |
निम्नलिखित के मान ज्ञात कीजिए : `cos^(-1)(cos""(3pi)/(6))` |
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Answer» Correct Answer - `pi/6` Range of `cos^(-1)` is `[0,pi]`. `cos^(-1)(cos(13pi)/6)=cos^(-1){cos(2pi+pi/6)}` `=cos^(-1){cospi/6}=pi/6` |
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| 23. |
Prove that i) `tan^(-1)(1+x)/(1-x)=pi/4 + tan^(-1)x,x lt 1` ii) `tan^(-1)x+cot^(-1)(x+1)=tan^(-1)(x^(2)+x+1)` |
| Answer» i) Put `x=tantheta`, ii) LHS `=tan^(-1)x+1/(tan^(-1)(x+1))` | |
| 24. |
i) Show that `sin^(-1){sin((3pi)/(4))} ne (3pi)/4` and find its value. ii) Show that `cos^(-1){cos(-pi/3)} ne -pi/3` and find its value. iii) Show that `tan^(-1){tan((5pi)/6)}ne (5pi)/6` and find its value. |
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Answer» i) We know that the principal-value branch of `sin^(-1)` is `[-pi/2,pi/2]`. `therefore sin^(-1){sin(3pi)/4} ne (3pi)/4`. Now, `sin^(-1){sin(3pi)/4)=sin^(-1){sin(pi-pi/4)}=sin^(-1){sinpi/4}[therefore sin(pi-pi/4)=sinpi/4]` `=pi/4 {therefore pi/4 in [-pi/2, pi/2]}`. `therefore sin^(-1){sin (3pi)/4}=pi/4`. ii) We know that the principal-value branch of `cos^(-1){cos(-pi/3)} ne -pi/3`. Now, `cos^(-1){cos(-pi/3)}=cos^(-1){cospi/3} [therefore cos(-theta)=costheta]=pi/3 [therefore pi/3 in [0,pi]]`. `therefore cos^(-1){cos(-pi/3)}=pi/3`. iii) We know that the principal-value branch of `tan^(-1)` is `(-pi/2,pi/2)`. `therefore tan^(-1){tan(5pi)/6} ne (5pi)/6`. Now, `tan^(-1)(tan(5pi)/6}=tan^(-1){tan(pi-pi/6)}=tan^(-1){tan(-pi/6)} [therefore tan(pi-theta)=tan(-theta)]`. `=-pi/6 [therefore -pi/6 in (-pi/2, pi/2)]` `therefore tan^(-1){tan(5pi)/6}=-pi/6`. |
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| 25. |
निम्नलिखित के मान ज्ञात कीजिए : `tan^(-1)(tan""(3pi)/(6))` |
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Answer» Correct Answer - `pi/6` Range of `tan^(-1)` is `(-pi/2,pi/2)` `therefore tan^(-1)(tan(7pi)/6)=tan^(-1){tan(pi+pi/6)}` `=tan^(-1){tanpi/6{=pi/6`. |
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| 26. |
Prove that `tan^(-1)((cosx)/(1+sinx))=(pi/4-x/2)` |
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Answer» We have, LHS `=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){tan(pi/4-x/2)}=(pi/4-x/2)`=RHS. `therefore tan^(-1)((cosx)/(1+sinx))=(pi/4-x/2)`. |
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| 27. |
Prove that `tan^(-1)((cosx-sinx)/(cosx+sinx))=(pi/4-x), x lt pi`. |
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Answer» We have, LHS `=tan^(-1)((cosx-sinx)/(cosx+sinx))` `=tan^(-1)((1-tanx)/(1+tanx))` [dividing num. and denom. By `cos x`] `=tan^(-1){tan(pi/4-x)}=(pi/4-x)` RHS. `therefore tan^(-1)((cosx-sinx)/(cosx+sinx))=(pi/4-x)`. |
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| 28. |
निम्नलिखित समीकरणों को सरल कीजिए : `2 tan^(-1)(cosx)=tan^(-1)(2 "cosec"x)` |
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Answer» We have, `2tan^(-1)(cosx)=tan^(-1)(2"cosec"x)` `rArr tan^(-1)(2cosx)/(1-cos^(2)x)=tan^(-1)(2"cosec"x)` `rArr tan[tan^(-1)(2cosx)/(sin^(2)x)]=2"cosec"x`. `rArr (2cosx)/(sin^(2)x)=2"cosec"x rArr cosx=sinx` `rArr tanx=1 rArr x=pi/4`. |
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| 29. |
`tan^(-1)""(cosx)/(1-sinx),-(-3pi)/(2) lt x lt (pi)/(2)` , को सरलता रूप में व्यक्त कीजिए । |
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Answer» We have, LHS `=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))/(2sin^(2)(pi/4-x/2)}` `tan^(-1){cot(pi/4-x/2)}=tan^(-1)[tan{pi/4-x/2)}]` `=tan^(-1){tan(pi/4+x/2)}` `=(pi/4+x/2)`=RHS. Hence, `tan^(-1)(cosx)/(1-sinx)=(pi/4+2/x)`. |
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| 30. |
`tan^(-1)""[(a cos x -b sinx)/(b cosx+a sinx)]` को सरल कीजिए, यदि `""(a)/(b)tanx le-1` |
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Answer» We have, LHS `=tan^(-1)(acosx-bsinx)/(bcosx-asinx)=tan^(-1){((acosx-bsinx)/(bcosx))/((bcosx-asinx)/(bcosx))` [on dividing num. and denom. By `bcosx`]. `=tan^(-1){(a/b-tanx)/(1+a/btanx)}=tan^(-1)(p-q)/(1+pq)`, where `a/b=p` and `tanx=q` `=tan^(-1)p-tan^(-1)q=tan^(-1)a/b-tan^(-1)(tanx)` `=(tan^(-1)a/b-x)`= RHS. Hence, `tan^(-1)((acosx-bsinx)/(bcosx-asinx))=(tan^(-1)a/b-x)`. |
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| 31. |
Prove that `tan^(-1)""(3a^(2)x-x^(3))/(a^(3)-3ax^(2))=3tan^(-1)""x/a`. |
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Answer» Putting `x=atantheta`, we get `tan^(-1)(3a^(2)x-x^(3))/(a^(3)-3ax^(2))=tan^(-1)((3a^(3)tantheta-a^(3)tan^(3)theta)/(a^(3)-3a^(3)tan^(2)theta))` `tan^(-1)(3tantheta-tan^(3)theta)/(1-3tan^(2)theta)=tan^(-1)(tan 3theta)` `=3theta=3tan^(-1)x/a`. Hence, `tan^(-1)(3a^(2)x-x^(3))/(a^(3)-3ax^(2))=3tan^(-1)x/a`. |
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| 32. |
`cot^(-1)(-1)` |
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Answer» Correct Answer - `(3pi)/4` Range of `cot^(-1)` is `(0,pi)` `therefore cot^(-1)(-1)=theta rArr cot theta=-1=-cotpi/4 = cot(pi-pi/4)=cot(3pi)/4 rArr theta=(3pi)/4`. |
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| 33. |
The principal value of `tan^-1(-sqrt3)` is |
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Answer» Correct Answer - `-pi/3` Range of `tan^(-1)` is `(-pi/2,pi/2)`. `therefore tan^(-1)(-sqrt(3))=theta rArr tantheta=-sqrt(3)=tan(-pi/3) rArr theta=-pi/3`. |
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| 34. |
`sec^(-1)(-2)` |
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Answer» Correct Answer - `(2pi)/3` Range of `sec^(-1)` is `[0,pi]-{pi/2}` `sec^(-1)(-2)=theta rArr sec theta=-2 =-secpi/3=sec(pi-pi/3)=sec((2pi)/3))`. `rArr theta=(2pi)/3`. |
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| 35. |
`tan^(-1)(-1)` |
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Answer» Correct Answer - `-pi/4` Range of `tan^(-1)` is `(-pi/2,pi/2)` `therefore tan^(-1)(-1)=theta rArr tan theta=-1=-tanpi/4=tan(-pi/4)` `rArr theta=-pi/4`. |
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| 36. |
`cos^(-1)(-1/2)` |
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Answer» Correct Answer - `(2pi)/3` Range of `cos^(-1)` is `[0,pi]` `therefore cos^(-1)(-1/2)=theta rArr cos theta=-1/2=-cospi/3=cos(pi-pi/3)=cos(2pi)/3` `rArr theta=(2pi)/3`. |
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| 37. |
The value of `sin(sin^(-1)(1/2)+cos^(-1)(1/2))=?` |
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Answer» Correct Answer - B `sin^(-1)1/2+cos^(-1)1/2=pi/2 rArr sin(sin^(-1)1/2+1/2+cos^(-1)1/2)=sinpi/2=1`. |
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| 38. |
Prove that `(9pi)/(8)-(9)/(4)sin^(-1)""(1)/(3)=(9)/(4)sin^(-1)""(2sqrt(2))/(3)` |
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Answer» We have, LHS `=(9pi)/8-9/4sin^(-1)1/3` `9/4(pi/2-sin^(-1)1/3)`. `=9/4cos^(-1)1/3=9/4sin^(-1)sqrt(1-1/9)` `9/4sin^(-1)(2sqrt(2))/(3)`=RHS Hence, LHS=RHS |
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| 39. |
The value of `sec^(-1)(sec(8pi)/5)` isA. `(2pi)/5`B. `(3pi)/5`C. `(8pi)/5`D. none of these |
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Answer» Correct Answer - A Let `sec^(-1)(sec(8pi)/5)=sec(2pi-(2pi)/5)=sec(2pi)/5 rArr x=(2pi)/5`. |
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| 40. |
The principal value of `cos^(-1)(-1/2)` isA. `-pi/3`B. `(2pi)/3`C. (4pi)/3`D. `pi/3` |
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Answer» Correct Answer - B Let `cos^(-1)(-1/2)=x`, where `x in [0,pi]`. Then, `cosx=-1/2=-cospi/3 = cos(pi-pi/3)=cos(2pi)/3 rArr x=(2pi)/3`. |
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| 41. |
Prove that: `cot^(-1)((ab+1)/(a-b))+cot^(-1)((bc+1)/(b-c))+cot^(-1)((ca+1)/(c-a))=0`. |
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Answer» We have, LHS `=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=RHS `therefore` LHS=RHS. |
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| 42. |
Prove that:`tan^(-1)(4/5)+cos^(-1)(12/13)=cos^(-1)(33/65)` |
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Answer» We have, LHS `=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-sqrt(1-(16/25)).sqrt(1-144/169)}` `=cos^(-1){48/65-sqrt(9/25).sqrt(25/169)}` `=cos^(-1){48/65-3/5 xx 5/13}=cos^(-1)(48/65-15/65)` `=cos^(-1){12/17-sqrt(64/289).sqrt(9/25)}=cos^(-1){12/17 -8/17 xx 3/5)` `=cos^(-1){12/17-24/85}=cos^(-1)(36/85)=`RHS Hence, `sin^(-1)8/17+sin^(-1)3/5=cos^(-1)36/85`. |
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| 43. |
`cos(tan^(-1)(3/4)) =`A. `3/5`B. `4/5`C. `4/9`D. none of these |
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Answer» Correct Answer - B Let `tan^(-1)3/4=x`, where `x in (-pi/2, pi/2)`. `therefore tanx=3/4` and since `x in (-pi/2, pi/2)`, we have `cosx gt 0`. `therefore cosx=1/(secx)=1/sqrt(1+tan^(2)x)=1/sqrt(1+9/16)=4/5`. `therefore cos(tan^(-1)3/4)=cosx=4/5`. |
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| 44. |
Range of `cos^(-1)x` isA. `[0,pi]`B. `[0,pi/2]`C. `[-pi/2,pi/2]`D. none of these |
| Answer» Correct Answer - A | |
| 45. |
Range of `sin^(-1)x` isA. `[0,pi/2]`B. `[0,pi]`C. `[-pi/2,pi/2]`D. none of these |
| Answer» Correct Answer - C | |
| 46. |
Prove that `2(tan^(-1)1/4+tan^(-1)2/9)=tan^(-1)4/3`. |
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Answer» We have, LHS =`2tan^(-1)1/4+2tan^(-1)2/9` `=tan^(-1)((2 xx 1/4)/(1-(1/4)^(2)) + tan^(-1)((2 xx 2/9)/(1-(2/9)^(2))` `=tan^(-1)((1/2)/(15/16))+tan^(-1)((4/9)/(77/81))` `=tan^(-1)(1/2 xx 16/15)+tan^(-1)(4/9xx 81/77)` `=tan^(-1)8/15+tan^(-1)36/77` `=tan^(-1)(1156)/867)=tan^(-1)4/3`=RHS `therefore` LHS=RHS. |
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| 47. |
If `tan^(-1) 3 +tan^(-1)x=tan^(-1)87` then x =A. `1/3`B. `1/5`C. 3D. 5 |
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Answer» Correct Answer - B `tan^(-1)x+tan^(-1)3=tan^(-1)8 rArr tan^(-1)(3+x)/(1-3x)=pi+tan^(-1)(-1)` `=pi-tan(1)=(pi-pi/4)=(3pi)/4`. |
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| 48. |
`(tan^(-1)2+tan^(-1)3)=?`A. `-pi/4`B. `pi/4`C. `(3pi)/4`D. `pi` |
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Answer» Correct Answer - C `(x=2,y=3) rArr xy gt 1`. `therefore tan^(-1)2+tan^(-1)3=pi+tan^(-1)(2+3)/(1-2 xx 3)=pi+tan^(-1)(-1)` `=pi-tan(1)=(pi-pi/4)=(3pi)/4`. |
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| 49. |
Find the principal value of: `cos^(-1)((sqrt(3))/2)`A. `pi/6`B. `(5pi)/6`C. `(7pi)/6`D. none of these |
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Answer» Correct Answer - A Let `cos^(-1)(sqrt(3)/2)=x,` where `x in [0,pi]` Then, `cosx=sqrt(3)/2=cospi/6 rArr x=pi/6` |
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| 50. |
If `tan^(-1)((x-2)/(x-4))+tan^(-1)((x+2)/(x+4))=pi/4`, find the value of `x`. |
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Answer» The given equation may be written as `tan^(-1)(x-2)/(x-4)=pi/4-tan^(-1)(x+2)/(x-4)=tan^(-1)1-tan^(-1)((x+2)/(x+4))[therefore pi/4=tan^(-1)(1/(x+2))`. `therefore (x-2)/(x-4) =1/(x+3) rArr (x-2)(x+3)=(x-4)` `rArr x^(2)+x-6=x-4 rArr x^(2)=2 rArr x=+-sqrt(2)`. Hence, `x=+-sqrt(2)`. |
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