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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.
| 51. |
Write the value of `sec^(2)theta(1+sin theta)(1-sin theta).` |
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Answer» Correct Answer - 1 Given expression `=sec^(2)theta(1-sin^(2)theta)=sec^(2)thetacos^(2)theta=1.` |
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| 52. |
If `x=a sin theta+b cos theta and y=a cos theta -b sin theta, " prove that " x^(2)+y^(2)=a^(2)+b^(2).` |
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Answer» We have ` x^(2) + y^(2) = (a sin theta + b cos theta)^(2) + (a cos theta - b sin theta)^(2) ` ` = a^(2)(sin^(2)theta + cos^(2)theta ) + b^(2)(cos^(2)theta + sin^(2)theta) ` `= a^(2) + b^(2) " " [ because sin^(2)theta + cos^(2)theta =1]. ` Hence, ` x^(2) + y^(2) = a^(2) + b^(2). ` |
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| 53. |
If ` sin theta + cos theta = m and sec theta + "cosec" theta = n, ` prove that ` n(m^(2)-1) = 2m. ` |
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Answer» We have ` n(m^(2) -1) = ( sec theta + "cosec" theta )[( sin theta + cos theta)^(2) -1] ` ` = ((1)/(cos theta) + (1)/(sin theta))[(sin^(2)theta + cos^(2)theta) + 2 sin theta cos theta -1 ] ` ` =((sin theta + cos theta))/(sin theta cos theta) * 2 sin theta cos theta ` `=2(sin theta + cos theta)= 2m. ` Hence, `n(m^(2) -1) = 2m. ` |
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| 54. |
Prove that `(i) (tan A+ sin A)/(tan A- sin A)= (sec A+1)/(sec A-1) ` ` (ii) (cot A- cos A)/(cot A+ cos A)= ("cosec" A-1)/("cosec" A+1) ` |
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Answer» We have `(i) " LHS " = (tan A + sin A)/(tan A - sin A)` ` = ((sin A)/(cos A)+ sin A)/((sin A)/(cos A)- sin A)=(sin A((1)/(cos A)+1))/(sin A((1)/(cos A)-1)) ` ` = (((1)/(cos A)+1))/(((1)/(cos A)-1))= (sec A+1)/(sec A-1)= " RHS. " ` ` therefore " LHS " = " RHS. " ` `(ii) " LHS " = (cot A - cos A)/(cot A + cos A) ` ` = ((cos A)/(sin A)- cos A)/((cos A)/(sin A)+ cos A)= (cos A((1)/(sin A)-1))/(cos A((1)/(sin A)+1))` ` =(((1)/(sin A)-1))/(((1)/(sin A)+1)) =(("cosec" A -1))/(("cosec" A +1))= "RHS. " ` ` therefore " LHS " = " RHS. " ` |
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| 55. |
Prove that `(sin theta - cos theta +1)/(sin theta + cos theta -1) = (1)/((sec theta - tan theta)). ` |
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Answer» LHS `=(sin theta - cos theta +1)/(sin theta + cos theta -1) ` ` = ((sin theta)/(cos theta)-1 + (1)/(cos theta ))/((sin theta )/(cos theta) +1 - (1)/(cos theta)) ` [ on dividing num. and denom. by ` cos theta ` ] ` =(tan theta -1+ sec theta )/(tan theta +1 - sec theta )=((sec theta + tan theta -1))/((tan theta - sec theta +1)) ` ` = ((sec theta + tan theta )-(sec^(2) theta - tan^(2) theta ))/((tan theta - sec theta +1)) " " [ because 1= sec^(2) theta - tan^(2) theta ] ` ` =((sec theta + tan theta )[1- (sec theta - tan theta )])/((tan theta - sec theta +1)) ` ` = ((sec theta + tan theta)(tan theta - sec theta +1))/((tan theta - sec theta +1))=(sec theta + tan theta). ` ` RHS= (1)/((sec theta - tan theta)) ` ` =(1)/((sec theta - tan theta))xx ((sec theta + tan theta ))/((sec theta + tan theta )) = ((sec theta + tan theta ))/((sec^(2)theta - tan^(2)theta))` `= (sec theta + tan theta) " "[ because sec^(2)theta - tan^(2)theta =1].` Hence,` LHS = RHS. ` |
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| 56. |
If `x=asinthetaa n dy=btantheta,`then prove that `(a^2)/(x^2)-(b^2)/(y^2)=1` |
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Answer» We have ` x= a sin theta rArr (a)/(x)=(1)/(sin theta) rArr (a)/(x) = "cosec" theta " "...(i) ` ` and y= b tan theta rArr (b)/(y)= (1)/(tan theta) rArr (b)/(y) = cot theta. " "...(ii) ` Squaring (i) and (ii) and subtracting, we get `((a^(2))/(x^(2))- (b^(2))/(y^(2)))= ( "cosec"^(2)theta - cot^(2)theta ) =1. ` Hence, ` ((a^(2))/(x^(2))- (b^(2))/(y^(2)))=1. ` |
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| 57. |
If `sec theta + tan theta = m, ` show that ` ((m^(2) -1))/((m^(2) +1)) = sin theta . ` |
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Answer» We have `(m^(2) -1)=(sec theta + tan theta)^(2) -1 ` ` = sec^(2) theta + tan^(2)theta + 2sec theta tan theta -1 ` ` = (sec^(2)theta -1) + tan^(2)theta + 2 sec theta tan theta ` `= 2 tan^(2)theta + 2sec theta tan theta " " [ because sec^(2)theta -1 = tan^(2)theta] ` ` = 2 tan theta(tan theta + sec theta). " "...(i)` ` (m^(2) +1) = (sec theta + tan theta)^(2) +1 ` ` = sec^(2) theta + tan^(2)theta + 2 sec theta tan theta +1 ` ` = (1+ tan^(2) theta)+ sec^(2)theta + 2sec theta tan theta ` `= 2 sec^(2) theta + 2 sec theta tan theta " " [ because 1+ tan^(2)theta = sec^(2)theta ] ` `=2 sec theta (sec theta + tan theta ). " "...(ii) ` From (i) and (ii), we get `((m^(2) -1))/((m^(2) +1)) = (tan theta)/(sec theta) = ((sin theta)/(cos theta ) xx cos theta) = sin theta . ` Hence, ` ((m^(2) -1))/((m^(2) +1)) = sin theta . ` |
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| 58. |
Write the value of ` "cosec"^(2)theta(1+costheta)(1-costheta).` |
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Answer» Correct Answer - 1 Given expression `"cosec"^(2)theta(1-cos^(2)theta)="cosec"^(2)thetasin^(2)theta=1.` |
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| 59. |
Prove that `(i) sec^(2) theta + "cosec"^(2) theta = sec^(2) theta "cosec"^(2) theta ` `(ii) tan^(2) theta - sin^(2)theta = tan^(2)theta sin^(2)theta ` `(iii) tan^(2) theta + cot^(2)theta +2 = sec^(2) theta "cosec"^(2) theta ` |
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Answer» We have ` (i) "LHS "= sec^(2)theta+ "cosec"^(2) theta ` `=(1)/(cos^(2)theta)+ (1)/(sin^(2)theta)=(sin^(2)theta+ cos^(2)theta)/(cos^(2)theta sin^(2)theta ) ` `= (1)/(cos^(2)theta sin^(2)theta) " "[ because sin^(2)theta + cos^(2) theta =1 ] ` ` = sec^(2)theta "cosec"^(2)theta = "RHS. " ` `therefore "LHS " = " RHS." ` `(ii) " LHS " = tan^(2)theta- sin^(2)theta ` ` =(sin^(2)theta)/(cos^(2)theta)- sin^(2)theta = (sin^(2)theta-sin^(2)theta cos^(2)theta )/(cos^(2)theta) ` ` =(sin^(2)theta(1- cos^(2)theta))/(cos^(2)theta)= (sin^(2)theta)/(cos^(2)theta)* sin^(2)theta ` ` = tan^(2)theta sin^(2)theta= "RHS." ` ` therefore "LHS " = " RHS. "` ` (iii) "LHS "= tan^(2)theta+ cot^(2)theta +2 ` ` = (1+tan^(2)theta)+(1+cot^(2)theta)=sec^(2)theta+ "cosec"^(2)theta ` ` =(1)/(cos^(2)theta)+(1)/(sin^(2)theta)=(sin^(2)theta+ cos^(2)theta)/(cos^(2)theta sin^(2)theta )` ` =(1)/(cos^(2)theta sin^(2)theta)= sec^(2)theta "cosec"^(2)theta = "RHS. "` ` therefore "LHS " = " RHS. " ` |
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| 60. |
Write the value of `sin^(2)theta cos^(2)theta(1+tan^(2)theta)(1+cot^(2)theta).` |
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Answer» Correct Answer - 1 Given expression `=sin^(2)theta(1+cot^(2)theta)cos^(2)theta(1+tan^(2)theta)` `=(sin^(2)theta "cosec"^(2)theta)(cos^(2)thetasec^(2)theta)=(1xx1)=1.` |
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| 61. |
Write the value of ` 3 cot^(2)theta - 3 "cosec"^(2)theta.` |
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Answer» Correct Answer - -3 Given expression `=3cot^(2)theta-3(1+cot^(2)theta)=-3.` |
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| 62. |
If `cosec theta - sin theta = a^3 and sec theta - cos theta = b^3` then prove that `a^2b^2(a^2+b^2) = 1.` |
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Answer» `a^(3)=((1)/(sin theta)-sin theta)=((1-sin^(2)theta)/(sintheta))=(cos^(2)theta)/(sintheta) rArr a=(cos^(2//3)theta)/(sin^(1//3)theta).` `b^(3)=((1)/(cos theta)-cos theta)=((1-cos^(2)theta)/(costheta))=(sin^(2)theta)/(costheta) rArr b=(sin^(2//3)theta)/(cos^(1//3)theta).` `therefore a^(2)b^(2)(a^(2)+b^(2))=a^(4)b^(2)+a^(2)b^(4)=a^(3)(ab^(2))+(a^(2)b)b^(3)` `=(cos^(2)theta)/(sintheta)*[(cos^(2//3)theta)/(sin^(1//3)theta)*(sin^(4//3)theta)/(cos^(2//3)theta)]+[(cos^(4//3)theta)/(sin^(2//3)theta)*(sin^(2//3)theta)/(cos^(1//3)theta)]*(sin^(2)theta)/(costheta)` `=(cos^(2)theta)/(sintheta)*sin theta+cos theta*(sin^(2)theta)/(cos theta)=(cos^(2)theta+sin^(2)theta)=1.` |
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| 63. |
If `cot theta+ tan theta=m` and `sec theta-cos theta=n` then `(m^2 n)^(2/3)-(mn^2)^(2/3)=` |
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Answer» `m=((costheta)/(sintheta)+(sintheta)/(costheta))=((cos^(2)theta+sin^(2)theta)/(sinthetacostheta))=(1)/(sinthetacostheta)` `n=((1)/(costheta)-costheta)=((1-cos^(2)theta)/(costheta))=(sin^(2)theta)/(costheta)` `therefore m^(2)n=((1)/(sin^(2)theta cos^(2)theta)xx(sin^(2)theta)/(cos^(2)theta))=(1)/(cos^(3)theta)=sec^(3)theta` `and mn^(2)=((1)/(sintheta costheta)xx(sin^(4)theta)/(cos^(2)theta))=(sin^(3)theta)/(cos^(3)theta)=tan^(3)theta.` `therefore (m^(2)n)^(2//3)-(mn^(2))^(2//3)=(sec^(3)theta)^(2//3)-(tan^(3)theta)^(2//3)=(sec^(2)theta-tan^(2)theta)=1.` |
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| 64. |
If `1+sin^2 theta= 3 sintheta cos theta,` then prove that `tan theta = 1` or `tan theta =1/2` |
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Answer» `1+sin^(2)theta=3sin thetacos theta rArr sec^(2)theta+tan^(2)theta=3 tan theta " "["dividing throughout by "cos^(2)theta]` `rArr (1+tan^(2)theta)+tan^(2)theta=3tantheta` `rArr 2tan^(2)theta-3tan theta+1=0` `rArr 2tan^(2)theta-2tantheta-tantheta+1=0` `rArr 2tantheta(tantheta-1)-(tan theta-1)=0` `rArr (tantheta-1)(2tantheta-1)=0` `rArr tan theta =1 or (1)/(2).` |
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| 65. |
Show that none of the following is an identity: ` (i) cos^(2) theta + cos theta =1 " " (ii) sin^(2) theta + sin theta =2 " " (iii) tan^(2) theta + sin theta = cos^(2) theta ` |
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Answer» `(i) cos^(2)30^(@)+cos 30^(@)=((sqrt(3))/(2))^(2)+(sqrt(3))/(2)=(3)/(4)+(sqrt(3))/(2)=(3+2sqrt(3))/(4) ne 1`. (ii) `sin^(2)30^(@)+sin30^(@)=((1)/(2))^(2)+(1)/(2)=(1)/(4)+(1)/(2)=(3)/(4) ne 2.` (iii) ` LHS=tan^(2)45^(@)+sin45^(@)=(1)^(2)+(1)/(sqrt(2))=1+(1)/(sqrt(2))=(sqrt(2)+1)/(sqrt(2))`. `RHS=cos^(2)45^(@)=((1)/(sqrt(2)))^(2)=(1)/(2).` `therefore LHS ne RHS.` |
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| 66. |
Prove: `(tanA+tanB)/(cotA+cotB)=tanAtanB` |
| Answer» `LHS=(tanA+tanB)/((1)/(tanA)+(1)/(tanB))=(tanAtanB(tanA+tanB))/((tanA+tanB))=tanAtanB.` | |
| 67. |
Prove that ` sqrt( sec^(2) theta + "cosec"^(2) theta )= tan theta + cot theta . ` |
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Answer» We have `LHS= sqrt( sec^(2) theta + "cosec"^(2) theta )` ` = sqrt((1+ tan^(2)theta)+(1+cot^(2)theta))= sqrt(tan^(2) theta + cot^(2)theta +2)` `= sqrt(tan^(2)theta + cot^(2)theta+ 2 tan theta * cot theta) " "[ because tan theta * cot theta =1] ` `=sqrt((tan theta + cot theta)^(2)) = tan theta + cot theta = RHS. ` ` therefore LHS = RHS. ` |
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| 68. |
Prove that `(i) (sin^(2)A cos^(2)B - cos^(2)A sin^(2) B )=(sin^(2)A- sin^(2)B) ` `(ii) (tan^(2)A sec^(2)B - sec^(2)A tan^(2)B)=(tan^(2)A- tan^(2)B) ` |
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Answer» We have `(i) LHS = (sin^(2)A cos^(2)B - cos^(2)A sin^(2) B )` `= sin^(2)A(1- sin^(2)B)-(1- sin^(2)A)sin^(2)B ` ` = sin^(2)A - sin^(2)B = RHS. ` `(ii) LHS = (tan^(2)A sec^(2)B - sec^(2)A tan^(2)B) ` ` =tan^(2)A(1+tan^(2)B) - (1+ tan^(2)A)tan^(2)B ` ` = (tan^(2)A - tan^(2)B) = RHS. ` |
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| 69. |
Prove the following identities:`tan^2A-tan^2B=(cos^2B-cos^2A)/(cos^2Bcos^2A)=(sin^2A-sin^2B)/(cos^2Acos^2B)``(sinA-sinB)/(cosA+cosB)+(cosA-cosB)/(sinA+sinB)=0` |
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Answer» We have ` (tan^(2)A - tan^(2)B) = (sin^(2)A)/(cos^(2)A) - (sin^(2)B)/(cos^(2)B) ` ` = (sin^(2)A cos^(2)B - cos^(2)A sin^(2)B)/(cos^(2)A cos^(2)B) ` ` = (sin^(2)A(1- sin^(2)B)-(1- sin^(2)A)sin^(2)B )/(cos^(2)A cos^(2)B) ` ` = ((sin^(2)A - sin^(2)B))/(cos^(2)A cos^(2)B).` Also, ` (tan^(2)A - tan^(2)B)= (sin^(2)A)/(cos^(2)A)- (sin^(2)B)/(cos^(2)B) ` ` = (sin^(2)A cos^(2)B - cos^(2)A sin^(2)B)/(cos^(2)A cos^(2)B) ` ` = ((1- cos^(2)A)cos^(2)B - cos^(2)A(1- cos^(2)B))/(cos^(2)B cos^(2)A) ` ` = ((cos^(2)B - cos^(2)A))/(cos^(2)B cos^(2)A). ` Hence, `(tan^(2)A - tan^(2)B) = ((sin^(2)A - sin^(2)B))/(cos^(2)A cos^(2)B) = ((cos^(2)B - cos^(2)A))/(cos^(2)B cos^(2)A). ` |
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| 70. |
If `xsin^3theta+ycos^3theta-sinthetacosthetaa n dxsintheta=ycostheta,`prove that `x^2+y^2=1` |
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Answer» ` xsin^(3) theta + y cos^(3) theta = sin theta cos theta ` ` rArr (xsin theta)sin^(2)theta + (ycos theta )cos^(2)theta = sin theta cos theta ` ` rArr (xsin theta)sin^(2)theta + (xsin theta)cos^(2)theta = sin theta cos theta " " [because y cos theta = x sin theta ] ` `rArr (xsin theta ) (sin^(2) theta + cos^(2) theta ) = sin theta cos theta ` ` rArr xsin theta = sin theta cos theta " " [ because sin^(2) theta + cos^(2) theta =1 ] ` ` rArr x= cos theta . " "...(i) ` Now ,`xsin theta = y cos theta ` `rArr cos theta sin theta = y cos theta " "[because x= cos theta] ` ` rArr y= sin theta. " "...(ii) ` On squaring (i) and (ii) and adding, we get ` x^(2) + y^(2) =1. ` Hence, ` x^(2) + y^(2) =1. ` |
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| 71. |
Choose the correct answer in each of the following questions: `sin^(2)30^(@)+4cot^(2)45^(@)-sec^(2)60^(@)=?` |
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Answer» Correct Answer - B `sin^(2)30^(@)+4cot^(2)45^(@)-sec^(2)60^(@)=((1)/(2))^(2)+4xx(1)^(2)-2^(2)=(1)/(4)+4-4=(1)/(4).` |
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