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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. |
If `sin ^(-1) .(2a)/(1+a^(2))+ coses^(2) (cot ^(-1) 3) y =?`A. `(a+b)/(1-ab)`B. `(a+b)/(1+ab)`C. `(a-b)/(1+ab)`D. None of these |
| Answer» Correct Answer - A | |
| 2. |
Prove that:`tan^(-1)1/5+tan^(-1)1/7+tan^(-1)1/3+tan^(-1)1/8=pi/4` |
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Answer» LHS=`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)""(11)/(23)` `=tan^(-1)""((6)/(17)+(11)/(23))/(1-(6)/(17)xx(11)/(23))=tan^(-1)""((138+187)/(391))/((391-66)/(391))` `tan^(-1)""(352)/(325)=tan^(-1)1=(pi)/(4)`= RHS Hence Proved |
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| 3. |
`If"sin"{cot^(-1)(x+1)}="cos"(tan^(-1)x),`then find `xdot`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))` |
| Answer» Correct Answer - D | |
| 4. |
If `(tan^(-1)x)^2+(cot^(-1)x)^2=(5pi^2)/8,`then find `xdot`A. 1B. -1C. `(1)/sqrt(3)`D. `-(1)/sqrt(3)` |
| Answer» Correct Answer - B | |
| 5. |
The Principal value of `os^(-1) (sqrt(3)/(2))` is :A. `(pi)/(8)`B. `(pi)/(6)`C. `(pi)/(3)`D. None of these |
| Answer» Correct Answer - B | |
| 6. |
If `sin(sin^(-1)1/5+cos^(-1)x)=1,`then find the value of `xdot` |
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Answer» `sin(sin^(-1)""(1)/(5)+cos^(-1)x)=1` `implies sin (sin^(-1)""(1)/(5)+cos^(-1)x)=sin""(pi)/(2)` `implies sin^(-1)""(1)/(5)+cos^(-1)x=(pi)/(2)` `implies sin^(-1)""(1)/(5)=(pi)/(2)-cos^(-1)x` `implies sin^(-1)""(1)/(5)=sin^(-1)ximplies x=(1)/(5)` |
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| 7. |
`sec^(2) (tan^(-1) 4) + cosec^(2) (cot ^(-1) 3) =?`A. 30B. 29C. 27D. 25 |
| Answer» Correct Answer - C | |
| 8. |
`tan^(-1)sqrt(3)-cot^(-1)(-sqrt(3))`is equal to(A) `pi` (B) `-pi/2` (C) 0 (D) `2sqrt(3)`A. `pi`B. `-(pi)/(2)`C. `0`D. `2sqrt(3)` |
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Answer» Correct Answer - B `tan^(-1)""sqrt(3)-cot^(-1)""(-sqrt(3))` `=tan^(-1)""sqrt(3)-(pi-cot^(-1)""sqrt(3))` `=tan^(-1)""sqrt(3)-pi+ cot^(-1)""sqrt(3)` `=(tan^(-1)""sqrt(3)+cot^(-1)""sqrt(3))-pi` `=(pi)/(2)-pi=-(pi)/(2)` |
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| 9. |
Prove that:`sin^(-1)8/(17)+sin^(-1)3/5=tan^(-1)(77)/(36)` |
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Answer» LHS=`sin^(-1)""(8)/(17)+sin^(-1)""(3)/(5)` `=sin^(-1)""[(8)/(17)sqrt(1-((3)/(4))^(2))+(3)/(5)sqrt(1-((8)/(17))^(2))]` `=sin^(-1)[(8)/(17)xx(4)/(5)+(3)/(5)xx(15)/(17)]` `( :.sin^(1)x=tan^(-1)""(x)/(sqrt(1-x^(2))))` `=sin^(-1)""(77)/(85)=tan^(-1)""(77//85)/(sqrt(1-((77)/(85))^(2)))` `tan^(-1)((77//875)/(36//85))=tan^(-1)""(77)/(36)`= RHS Hence proved. |
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| 10. |
Find the value of: `tan1/2[sin^(-1)(2x)/(1+x^2)+cos^(-1)(1-y^2)/(1+y^2)],|x|0`and `x y" " |
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Answer» `tan""(1)/(2)[sin^(-1)""(2x)/(1+x^(2))+cos^(-1)""(1-y^(2))/(1+y^(2))]` `=tan{(1)/(2)(2 tan^(-1)x+2 tan^(-1)y)}" " { :. 2 tan^(-1)=sin^(-1)""(2x)/(1+x^(2)) and 2 tan^(-1)y=cos^(-1)""(1-y^(2))/(1-xy)}` `=tan(tan^(-1)x+tan^(-1)y)` `=tan(tan^(-1)""(x+y)/(1-xy))=(x+y)/(1-xy)` |
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| 11. |
Find the value of: `cot(tan^(-1)a+cot^(-1)a)` |
| Answer» `cot(tan^(-1)a+cot^(-1)a)=cot""(pi)/(2)=0` | |
| 12. |
Find the principal valuesof each of the following:`cot^(-1)(-sqrt(3))`(ii) `cot^(-1)(sqrt(3))` |
| Answer» `cot^(-1)(sqrt(3))=cot^(-1)(cot""(pi)/(6))=(pi)/(6)` | |
| 13. |
`tan^(-1)""(x)/(sqrt(a^(2)-x^(2))),|x|lt a` |
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Answer» `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)/(sqrt(a^(2)(1-sin^(2)theta)))" " Let x=a sin theta implies sin theta =(x)/(a)implies theta=sin^(-1)""((x)/(a))` `=tan^(-1)""(a sin theta)/(sqrt(a^(2)cos^(2)theta))` `=tan^(-1)""(a sin theta)/(a cos theta)=tan^(-1)(tan theta)` `=theta=sin^(-1)""(x)/(a)` |
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| 14. |
Prove that: `3cos^(-1)x=cos^(-1)(4x^3-3x), x in [1/2,1]` |
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Answer» Let `cos^(-1)x=theta` `RHS=cos^(-1)(4x^(3)-3x)` `=cos^(-1)(4cos^(3)theta- 3 cos theta)` `= cos^(-1)(cos 3 theta)` `=3theta=3 cos^(-1)x` = LHS Hence Proved. |
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| 15. |
Prove that `cos ^(-1) x = 2 sin ^(-1).sqrt(1-x)/(2)` |
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Answer» Let `cos^(-1) x = theta` `rArr" "x - cos theta` R.H.S `= 2 sin sqrt((1-x)/(2)) = 2sin. sqrt((1-costheta))/(2)` `= 2 sin ^(-1).sqrt((2 sin^(2)((theta)/(2)))/(2)) = 2 sin (sin.(theta)/(2)`) ` 2.(theta)/(2)= theta= cos^(-1) x= L.H.S` |
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| 16. |
If `sin^(-1)x=y`,then :A. `o le y le pi `B. `-(pi)/(2) le y le (pi)/(2)`C. `o lt y lt pi `D. `-(pi)/(2) lt y lt (pi)/(2)` |
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Answer» Correct Answer - B We known that the range of `sin^(-1)x is [-(pi)/(2),(pi)/(2)]` ` :. -(pi)/(2) le y le (pi)/(2)` |
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| 17. |
Prove that: `3sin^(-1)x=sin^(-1)(3x-4x^3), x in [-1/2,1/2]` |
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Answer» Let `sin^(-1)x=theta` ` implies x =sin theta` RHS`=sin^(-1)(3x-4x^(3))` `=sin^(-1)(sin3theta)` `= 3 theta` `= 3 sin^(-1)x`= LHS. Hence Proved. |
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| 18. |
`tan^-1 (x-1)/(x-2) + tan^-1 (x+1)/(x+2)=pi/4.` find |
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Answer» `tan^(-1)""(x-1)/(x-2)+tan^(-1)""(x+1)/(x+2)=(pi)/(4)` `implies tan^(-1)""((x-1)/(x-2)+(x+1)/(x+2))/(1-(x-1)/(x-2)*(x+1)/(x+2))=tan^(-1)1` `((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)=1implies 2x^(2)-4= -3` `implies 2x^(2)=1 implies x^(2)=(1)/(2)impliesx=pm(1)/(sqrt(2))` |
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| 19. |
Write the value of `cos^(-1)(1/2)+2 sin^(-1)(1/2)` |
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Answer» `cos^(-1)((1)/(2))+ 2 sin^(-1) ""(1)/(2)` `=cos^(-1)(cos""(pi)/(3))+2sin^(-1)(sin""(pi)/(6))` `(pi)/(3)+2((pi)/(6))=(pi)/(3)+(pi)/(3)=(2pi)/(3)` |
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| 20. |
Prove that `tan^(-1) x + cot^(-1) (x+1) = tan ^(-1) (x^(2) + x+1)`. |
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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) xx(2)/(9))` `= tan^(-1).(x+(1)/(x+1))/(1-x.(1)/(x+1))= tan ^(-1).(x(x-1)+1)/((x+ 1)-x)` `tan^(-1) (x^(2) + x+1)` = R.H.S |
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| 21. |
Prove that: `tan^-1(1/4)+tan^-1(2/9)=1/2 cos^-1(3/5)`. |
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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)xx(2)/(9))` ` tan^(-1). (9+8)/(36-2) = tan^(-1).(1)/(2) = (1)/(2) .2 tan ^(-1).(1)/(2)` ` = (1)/(2)cos^(-1).(1-(1)/(2^(2)))/(1+(1)/(2^(2))) (because cos ^(-1) .(1 -x^(2))/(1+ x^(2)) = 2 tan^(-1)x)` ` (1)/(2) cos^(-1).(3//4)/(5//4) = (1)/(2) cos^(-1).(3)/(5) = R.H.S` |
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| 22. |
Find the principalvalues of `cos^(-1)(sqrt(3))/2`and `cos^(-1)(-1/2)` |
| Answer» `cos^(-1)(-(1)/(2))=pi-cos^(-1)((1)/(2))=pi-cos^(-1)(cos""(pi)/(3))=pi-(pi)/(3)=(2pi)/(3)` | |
| 23. |
Prove that ` tan (2 tan^(-1) x ) = 2 tan (tan^(-1) x + tan^(-1) x^(3))`. |
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Answer» L.H.S `= tan(2 tan^(-1)x) = tan{ tan^(-1) .(2x)/(1-x^(2))}=(2x)/(1-x^(2))` `R.H.S = 2 tan {tan ^(-1) x+ tan ^(-1)x^(3)}` ltbr. `= 2 tan { tan^(-1) .(x+x^(3))/(1-x.x^(3))} = (2x(1+x^(2)))/((1-x^(2))(1+x^(2)))=(2x)/(1-x^(2))` L.H.S= R.H.S |
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| 24. |
Evaluate ` 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)/(2))/(1-(1)/(2) xx(1)/(3)) = tan^(-1).(x(x+1)+1)/((x+1)-x)` `=tan^(-1) (x^(2) +x^+1)` = R.H.S |
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| 25. |
If `tan^(- 1)x+tan^(- 1)y+tan^(- 1)z=pi` prove that `x+y+z=xyz` |
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Answer» `tan^(-1) x + tan ^(-1) y + tan ^(-1)z =pi ` `rArr " "tan^(-1).(x+y)/(1-xy) + tan^(-1) z = pi` `tan^(-1) .((x+y)/(1-xy))/(1-(x+y)/(1-xy)z)` `rArr (x+y +z(1-xy))/(1-xy -(x+y)z)= tan pi =0` `rArr " m " x+y + z -xyz =0` `rArr" "x +y + z =xyz` |
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| 26. |
`tan^(-1)(tan (3pi)/(4))` |
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Answer» `tan^(-2)tan((3pi)/(4))=tan^(-2)tan(pi-(pi)/(4)) " " [ because " Range of " tan^(-1)x=(-(pi)(2),(pi)/(2))]` `tan^(-1)(-tan""(pi)/(4))` `=-tan^(-1)(tan""(pi)/(4))=-(pi)/(4)` |
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| 27. |
`cos^(-1)(-sqrt(2))` |
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Answer» `cos^(-1)(-(1)/(sqrt(2)))=pi-cos^(-1)((1)/(sqrt(2)))=pi-cos^(-1)(cos""(pi)/(4))` `pi-(pi)/(4)=(3pi)/(4)` |
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| 28. |
Find the value of: `tan^(-1)(1)+cos^(-1)(-1/2)+sin^(-1)(-1/2)` |
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Answer» `tan^(-1)(1)+cos^(-1)(-(1)/(2))+sin^(-1)(-(1)/(2)) " " ( :. sin^(-1)x+cos^(-1)x=(pi)/(2))` `=tan^(-1)(tan""(pi)/(4))+(pi)/(2)=(pi)/(4)+(pi)/(2)=(3pi)/(4)` |
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| 29. |
Find the value of `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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| 30. |
`sin(pi/3-sin^(-1)(-1/2))`is equal to(A) `1/2` (B) `1/3` (C) `1/4` (D) 1A. `(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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| 31. |
If `tan^(-1)x = theta`, find the value of `sin^(-1).(2x)/(1+x^(2))` |
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Answer» `tan^(-1) x = theta` `rArr" "x tan = theta` Now `sin ^(-1).(2x)/(1+x^(2)) = sin ^(-1) .(2tan theta)/(1+ tan^(2)theta)` `" "= sin ^(-1)(sin 2theta)` `" " = 2theta` |
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| 32. |
Find the principalvalue of `tan^(-1)(-sqrt(3))` |
| Answer» `tan^(-1)(-sqrt(3))=-tan^(-1)(sqrt(3)) = - tan^(-1)(tan""(pi)/(3))=-(pi)/(3)` | |
| 33. |
`theta=s ec^(- 1)(2/(sqrt(3)))` |
| Answer» `sec^(-1)""((2)/(sqrt(3)))=sec^(-1)(sec""(pi)/(6))=(pi)/(6)` | |
| 34. |
Prove that `sec^(2) (tan ^(-1) 3) + cosec^(2)(cot^(-1)2) = 15` |
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Answer» Let `tan^(-1) 3 = A and cot^(-1) 2 = B` `rArr" " tanA = 3 and cot B = 2 ` `L.H.S = sec^(2) (tan ^(-1)3)+ cosec^(2) (cot^(-1)2)` `= sec^(2) A + cosec^(2) (cot^(-1) 2)` `= sec^(2)A + cosec^(2) B` `= 1 + tan^(2)A + 1 + cot ^(2)B` `2+3 ^(2)+2^(2) = 15 = R.H.S` |
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| 35. |
`cosec"^(-1)(-sqrt(2))` |
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Answer» `"cosec"^(-1)(-sqrt(2))=-cosec^(-1)(sqrt(2))` `=- "cosec"^(-1)("cosec"(pi)/(4))=-(pi)/(4)` |
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| 36. |
Find: `sin^(-1)(sin (2pi)/(3))` |
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Answer» `sin(sin""(2pi)/(3))=sin^(-1)[sin(pi-(pi)/(3))] " " { :. " Range of " sin^(-1)x=[-(pi)/(2),(pi)/(2)]}` `sin^(-1)(sin""(pi)/(3))=(pi)/(3)` |
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| 37. |
Prove that ` sin ^(-1).(3)/(5) = tan ^(-1) .(3)/(4)`. |
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Answer» Let `sin ^(-1).(3)/(5) =theta` `rArr" "sin theta =(3)/(5) rArr cosec theta=(5)/(3)` `rArr " "cosec^(2) theta =(25)/(9) " "rArr 1 + cot^(2) theta=(25)/(9)` `rArr" "cot^(2)theta =(16)/(9) rArr " "cot theta =(4)/(3)` `rArr" "sin ^(-1).(3)/(5) = tan^(-1).(3)/(4)` |
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| 38. |
Find the value of the following: `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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| 39. |
`cos^(-1)(cos((7pi)/6))` is equal toA. `(7pi)/(6)`B. `(5pi)/(6)`C. `(pi)/(3)`D. `(pi)/(6)` |
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Answer» Correct Answer - B `cos^(-1)""(cos""(7pi)/(6))=cos^(-1)""[cos(2pi-(5pi)/(6))]` `:.` Range of `cos^(-1)x=[0,pi]` `=cos^(-1)(cos""(pi)/(6))=(5pi)/(6)` |
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| 40. |
If `sin^(-1)(1-x)-2 sin^(-1)x=(pi)/(2)`, then `x` is equal toA. `0,(1)/(2)`B. `1,(1)/(2)`C. `0`D. `(1)/(2)` |
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Answer» Correct Answer - C `sin^(-1)(1-x)-2sin^(-1)x=(pi)/(2)" " Let sin^(-1)x=theta` `" "implies x=sintheta` `implies sin^(-1)(1-sintheta)-2theta=(pi)/(2)` `impliessin^(-1)(1-sin theta)=((pi)/(2)+2 theta)` `1-sin theta=sin((pi)/(2)+2theta)` `= cos 2 theta=1-sin^(2)theta` `implies 2sin^(2) theta -sin theta =0 implies 2x^(2)-x=0` `implies x(2x-1)=0 impliesx=0 or x=(1)/(2)` but the given equation is not satisfied by x`=(1)/(2)` `: . x=0` |
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| 41. |
`tan^(-1)sqrt(3)-sec^(-1)(-2)`is equal to(a) `pi` (B) `-pi/3`(C) `pi/3` (D) `(2pi)/3`A. `pi`B. `-(pi)/(3)`C. `(pi)/(3)`D. `(2pi)/(3)` |
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Answer» `tan^(-1)sqrt(3)-sec^(-1)(-2)` `=tan^(-1)(tan""(pi)/(3))-[pi-sec^(-1)2]` `=(pi)/(3)-pi+sec^(-1)(sec""(pi)/(3))` `=(pi)/(3)-pi+(pi)/(3)=-(pi)/(3)` |
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| 42. |
If `cos ^(-1) .(x)/(a)+ cos ^(-1). (y)/(b) = theta` then prove that `(x^(2))/(a^(2)) -(2xy)/(ab) . cos theta+ (y^(2))/(b^(2)) = sin ^(2) theta`. |
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Answer» `cos^(-1).(x)/(a) + cos theta +(y^(2))/(b^(2)) = sin^(2) theta` `rArr cos^(-1)[(x)/(a).(y)/(b)-sqrt(1-(x^(2))/(a^(2)))sqrt(1-(y^(2))/(b^(2)))] = theta` `rArr (xy)/(ab) -sqrt(1-(x^(2))/(a^(2))-(y^(2))/(b^(2))+(x^(2)y^(2))/(a^(2)b^(2)))= cos theta` `rArr " " (xy)/(ab)-costheta =sqrt(1-(x^(2))/(a^(2))-(y^(2))/(b^(2))+(x^(3)y^(2))/(a^(2)b^(2)))` squaring both sideswe get `(x^(2)y^(2))/(a^(2)b^(2))+ cos^(2) theta -(2xy)/(ab) cos theta = 1- (x^(2))/(a^(2))-(y^(2))/(b^(2))+(x^(2)y^(2))/(a^(2)b^(2))` `rArr (x^(2))/(a^(2)) -(2xy)/(ab) cos theta +(y^(2))/(b^(2)) = 1cos^(2) theta` `rArr (x^(2))/(y^(2)) -(2xy)/(ab) costheta +(y^(2))/(b^(2)) = sin^(2)theta` Hence Proved. |
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| 43. |
`sin^(-1) (sin=(2pi)/(3)) = ?`A. `(pi)/(3)`B. `(pi)/(4)`C. `(2pi)/(3)`D. None of these |
| Answer» Correct Answer - A | |
| 44. |
`"cosec"^(-1)(2)` |
| Answer» `"cosec"^(-1)("cosec"(pi)/(6))=(pi)/(6)` | |
| 45. |
`tan^(-1).(x(x+1)+1)/((x+1)-x)` `=tan^(-1) (x^(2) +x^+1)` = R.H.S |
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Answer» `sin.(7pi)/(4) = sin (2pi -(pi)/(4))= -sin .(pi)/(4)= - (1)/sqrt(2)` `therefore sin^(-1)(sin .(7pi)/(4)) = sin^(-1) (sin.(pi)/(4)) = - (pi)/(4)` |
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| 46. |
Evaluate `cot (tan ^(-1) 3)`. |
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Answer» Let `tan^(-1)3 = x` ` rArr" "tan x = 3` `rArr" "cot x =(1)/(3)` `rArr" "cot(tan^(-1)3) =(1)/(3)` |
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| 47. |
Find the value of the following: `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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| 48. |
If `sin^(-1)x + tan ^(-1) x = (pi)/(2)`, then prove that `2x^(2) + 1 = sqrt(5)` |
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Answer» ` sin^(-1) x + tan ^(-1) x =(pi)/(2)` ` rArr tan^(-1) x =(pi)/(2)- sin^(-1) x` `" "=cos^(-1)x = tan^(-1).(sqrt(1-x^(2)))/(x)` `rArr " "x=sqrt(1-x^(2))/(x)` `rArr" " x^(2) = sqrt(1-x^(2))` `" "x^(4)+x^(2) -1= 0` `rArr " "x^(2)=(-1pmsqrt(1+4))/(2xx1)` `rArr " "2x^(2) = -1 pm sqrt(5)` `rArr " " 2x^(2) + 1 = sqrt(5)` `" "(because 2x^(2)+ 1 = - sqrt(5)" is not possible")` |
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| 49. |
` cos^-1[sqrt3/2]` |
| Answer» `cos^(-1)""((sqrt(3))/(2))= cos^(-1)""(cos""(pi)/(6))=(pi)/(6)` | |
| 50. |
If `sin ^(-1) x =(pi)/(4)`, find the value of `cos^(-1) x`. |
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Answer» `sin^(-1)x =(pi)/(4)` ` x = sin.(pi)/(4) =(1)/sqrt(2)` `cos^(-1) x = cos^(-1) (1)/sqrt(2)` `cos ^(-1) (cos.(pi)/(4)) =(pi)/(4)` |
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