InterviewSolution
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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. |
Let `f : (-1, 1) -> R` be such that `f(cos4theta) = 2/(2-sec^2theta` for `theta in (0, pi/4) uu (pi/4, pi/2)`. Then the value(s) of `f(1/3)` is/areA. `1-sqrt((3)/(2))`B. `1+sqrt((3)/(2))`C. `1-sqrt((2)/(3))`D. `1+sqrt((2)/(3))` |
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Answer» Correct Answer - A::B `f(cos 4theta) = (2)/(2-sec^(2) theta)" ...(i)"` At `cos 4 theta = (1)/(3)` `implies 2cos^(2)2 theta-1= (1)/(3)` `impliescos^(2)2 theta=(2)/(3)` `implies cos2 theta = pmsqrt((2)/(3)) " ...(ii)"` `therefore f(cos 4 theta) = (2 cos^(2)theta)/(2 cos^(2)theta-1)` `=(1+cos2 theta)/(cos 2 theta)` `implies f((1)/(3)) = 1 pm sqrt((3)/(2)) " [from Eq. (ii)]"` |
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| 2. |
Given `alpha+beta+gamma=pi,`prove that `sin^2alpha+sin^2beta-sin^2gamma=2sinalphasinbetacosgammadot` |
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Answer» `LHS = sin^(2)alpha+sin^(2)beta-sin^(2)gamma` `sin^(2)alpha+(sin^(2)beta-sin^(2)gamma)` `=sin^(2)alpha + sin(beta+gamma)sin(beta-gamma)` `=sin^(2)alpha + sin (pi -alpha)sin(beta - gamma)[because alpha+beta+gamma = pi]` `=sin^(2)alpha+sinalphasin(beta-gamma)` `=sin alpha [sinalpha + sin(beta-gamma)]` `=sin alpha[sin (pi-(beta+gamma))+sin(beta-gamma)]` `=sinalpha[sin(beta+gamma)+sin(beta-gamma)]` `=sin alpha[2 sin beta cos gamma]` `=2 sin alpha sin beta cos gamma = RHS` |
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| 3. |
If A + B + C = `180^(@)`, then prove that tan A + tan B + tan C = tan A tan B tan C. |
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Answer» Since, A + B = `180^(@)C` `therefore tan(A + B) = tan(180^(@)-C)` `implies (tanA+tanB)/(1-tanAtanB)=-tanC` `implies tanA + tanB =- tan C + tan A tan B tan C` `implies tanA + tanB +tanC = tanA tanB tanC` |
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| 4. |
prove that : `tan(alpha)+2 tan(2alpha) +4(tan4alpha)+8cot(8alpha) = cot(alpha)` |
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Answer» We know that, `cot theta-tantheta=(1-tan^(2)theta)/(tan theta)=2((1-tan^(2)theta)/(2tantheta))=2cot 2 theta...(i)` LHS = `tan alpha + 2 tan 2 alpha + 4 tan 4 alpha + 8 cot 8 alpha` `=-(cotalpha - tan alpha - 2 tan 2 alpha - 4tan 4 alpha)+8cot 8 alpha + cot alpha` `=-(2cot 2alpha-2tan 2 alpha - 4tan 4 alpha)+8cot 8alpha + cotalpha" [from Eq. (i)]"` `=-(2(2cot 4 alpha)-4 tan 4 alpha)+8cot 8 alpha+cot alpha " [from Eq. (i)]"` `=-4(cot4 alpha-tan 4alpha)+8cot8alpha + cot alpha` `=-8cot 8 alpha +8cot 8 alpha + cot alpha " [from Eq. (i)]"` `=cot alpha = RHS` |
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| 5. |
If tan A = `(1-cos B)//sinB`, then tan 2A = tan B. |
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Answer» Correct Answer - True Since, `tanA = (1-cosB)/(sinB) = (2sin^(2).(B)/(2))/(2sin.(B)/(2)cos.(B)/(2))` `tanA = tan B//2` `implies tan 2A = tan B` Hence, it is a true statement. |
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| 6. |
In any triangle ABC prove that `cot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)cot(B/2)cot(C/2)` |
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Answer» Since, A + B + C = `pi` `implies (A)/(2) + (B)/(2) = (pi)/(2) - (C)/(2)` `implies cot ((A)/(2)+(B)/(2))=cot((pi)/(2)-(C)/(2))` `implies (cot.(A)/(2).cot.(B)/(2)-1)/(cot.(B)/(2)+cot.(A)/(2))=tan.(C)/(2)` `implies cot.(A)/(2).cot.(B)/(2).cot.(C)/(2)-cot.(C)/(2)=cot.(A)/(2)+cot.(B)/(2)` `implies cot.(A)/(2)+cot.(B)/(2)+cot.(C)/(2)=cot.(A)/(2)cot.(B)/(2)cot.(C)/(2)` |
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| 7. |
If the lengths of the sides of a triangle are in AP and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle isA. `3 : 4 : 5`B. `4 : 5 : 6`C. `5 : 9 : 13`D. `5 : 6 : 7` |
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Answer» Correct Answer - B Let a, b and c be the lengths of sides of a `DeltaABC` such that `a lt b lt c`. Since, sides are in AP. `therefore 2b = a + c " ...(i) "` Let `angleA = theta` Then, `angleC = 2 theta " [according to the equation]"` So, `angleB = pi - 3 theta " (ii)"` On applying sine rule in Eq. (i), we get 2 sin B = sin A + sin C `implies 2 sin (pi - 3 theta) = sin theta + sin 2 theta " [from Eq. (ii)]"` `implies 2 sin 3 theta = sin theta + sin 2 theta` `implies 2 [3sin theta - 4 sin^(3) theta]= sin theta + 2sin theta cos theta` `implies6 - 8sin^(2) theta = 1 + 2 cos theta" "[because sin theta " can not be zero]"` `implies 6 - 8(1 - cos^(2)theta) = 1 + 2cos theta` `implies 8 cos^(2) theta - 2 cos theta - 3 = 0` `implies (2 cos theta + 1)(4 cos theta - 3) = 0` `implies cos theta = (3)/(4)` or `cos theta = - (1)/(2)` (rejected). Clearly, the ratio of sides is a : b : c `=sin theta : sin 3 theta : sin 2 theta` `=sin theta : (3sin theta - 4 sin^(3) theta) : 2 sin theta cos theta` `=1 : (4 cos^(2) theta - 1) : 2 cos theta` `=1 : (5)/(4):(6)/(4) = 4 : 5: 6` |
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| 8. |
If `A >0,B >0a n dA+B=pi/3,`the maximum value of `tanAtanB`is______ |
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Answer» Correct Answer - `(1)/(3)` Since, A+b `=(pi)/(3)` and, we know product of term is maximum, when values are equal. `therefore(tan A. tan B)` is maximum. When A = B = `pi//6` i.e. `y = tan.(pi)/(6)tan.(pi)/(6)=(1)/(3)` |
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| 9. |
The value of sin `10^(@)` sin `30^(@)` sin `50^(@)` sin `70^(@)` isA. `(1)/(36)`B. `(1)/(32)`C. `(1)/(16)`D. `(1)/(18)` |
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Answer» Correct Answer - C We have, sin `10^(@) sin 50^(@) sin 70^(@)` `=sin (30^(@))[sin(10^(@))sin(50^(@))sin (70^(@))]` `=(1)/(2)[sin(10^(@))sin(60^(@)-10^(@))sin(60^(@)+10^(@))]` `=(1)/(2)[(1)/(4)sin(3(10^(@)))] " "[because sin theta sin (60^(@)-theta)sin(60^(@)+theta)=(1)/(4)sin3 theta]` `=(1)/(8)sin 30^(@) = (1)/(8) xx (1)/(2) = (1)/(16)` |
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| 10. |
If `A = sin^2 theta+ cos^4 theta`, then for all real values of `theta`A. `1leAle2`B. `(3)/(4) le A le 1`C. `(13)/(16) le A le 1`D. `(3)/(4) le A le (13)/(16)` |
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Answer» Correct Answer - B Given, `A - sin^(2) theta+(1-sin^(2) theta)^(2)` `implies A = sin^(4) theta - sin^(2) theta + 1` `implies A = (sin^(2)theta-(1)/(2))^(2)+(3)/(4)` `implies 0le (sin^(2)theta-(1)/(2))^(2)le(1)/(4) " "[because 0 le sin^(2) theta le 1]` `therefore (3)/(4) le A le 1` |
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| 11. |
`(1+cos.(pi)/(8))(1+cos.(3pi)/(8))(1+cos.(5pi)/(8))(1+cos.(7pi)/(8))` is equal toA. `(1)/(2)`B. `cos.(pi)/(8)`C. `(1)/(8)`D. `(1+sqrt(2))/(2sqrt(2))` |
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Answer» Correct Answer - C Given expression = `(1+cos.(pi)/(8))(1+cos.(3pi)/(8))(1+cos.(5pi)/(8))(1+cos.(7pi)/(8))` `(1+cos.(pi)/(8))(1+cos.(3pi)/(8))(1-cos.(3pi)/(8))(1-cos.(pi)/(8))` `=(1-cos^(2).(pi)/(8))(1-cos^(2).(3pi)/(8))` `=(1)/(4)(2-1-cos.(pi)/(4))(2-1-cos3.(pi)/(4))` `=(1)/(4)(1-cos.(pi)/(4))(1-cos3.(pi)/(4))` `=(1)/(4)(1-(1)/(sqrt(2)))(1+(1)/(sqrt(2)))=(1)/(4)(1-(1)/(2))=(1)/(8)` |
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| 12. |
Which of the following numbers is rational?A. sin `15^(@)`B. `cos 15^(@)`C. sin `15^(@)` cos `15^(@)`D. sin `15^(@)` cos `75^(@)` |
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Answer» Correct Answer - C Since, `sin 15^(@) = (1)/(2) sqrt(2-sqrt(3)) and cos 15^(@) = (1)/(2) sqrt(2+sqrt(3))` and `sin 15^(@) cos 75^(@) = sin 15^(@).sin 15^(@) = (1)/(4)(2 - sqrt(3))` Therefore, all these values are irrational and `sin 15^(@) cos 15^(@) = (1)/(2).2 sin 15^(@) cos 15^(@)` `=(1)/(2).sin 30^(@) = (1)/(4)`, which is rational. |
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| 13. |
Prove that: `3 (sin x-cos x)^4+ 6 (sin x +cosx)^ 2+4 (sin^6 x+ cos^6 x) -13=0`A. 11B. 12C. 13D. 14 |
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Answer» Correct Answer - C Given expression = `3(sinx-cosx)^(4) + 6(sinx + cosx)^(2) + 4(sin^(6)x + cos^(6)x)` `=3(1-sin2x)^(2) + 6(1 + sin2x) + 4{(sin^(2) x+cos^(2)x)^(3) - 3 sin^(2)x cos^(2)x(sin^(2)x+cos^(2)x)}` `=3(1-2sin 2x + sin^(2)2x)+6+6sin2x+4(1-3sin^(2)x cos^(2)x)` `=3(1-2 sin 2x + sin^(2)2x + 2 +2sin2x)+4(1-(3)/(4).sin^(2)2x)` `=13+3sin^(2)2x - 3 sin^(2)2x =13` |
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| 14. |
Find the value of the expression `3[sin^(4)((3pi)/(2)-alpha)+sin^(4)(3pi+alpha)]-2[sin^(6)((pi)/(2)+alpha)+sin^(6)(5pi-alpha)]`. |
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Answer» Correct Answer - B Given expression = `3[sin^(4)((3pi)/(2)-alpha)+sin^(4)(3pi+alpha)]-2[sin^(6)((pi)/(2)+alpha)+sin^(6)(5pi-alpha)]` `=3(cos^(4)alpha + sin^(4) alpha)-2(cos^(6) alpha + sin^(6) alpha)` `=3(1-2sin^(2) alpha cos^(2) alpha)-2(1-3sin^(2) alpha cos^(2) alpha)` `=3-6 sin^(2) alpha cos^(2) alpha - 2+6 sin^(2) alpha cos^(2) alpha =1 ` |
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| 15. |
Match the conditions/expressions in Column I with values in Column II `(sin 3 alpha)//cos2 alpha)` is `{:(" Column I"," Column II"),("A. Positive","p."(13pi//48, 14pi//48)),("B. negative","q."(14pi//48, 18pi//48)),(,"r."(18pi//48, 23pi//48)),(,"s."(0, pi//2)):}` |
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Answer» Correct Answer - A `rarr` r; B `rarr` p In the interval `((13pi)/(48),(14pi)/(48)),cos2 alpha lt 0 and sin 3 alpha gt 0`. `implies(sin3alpha)/(cos2alpha)` is negative, therefore B `rarr` p. Again in the interval `((18pi)/(48),(23pi)/(48))`, both `sin 2 alpha` and `cos 2 alpha` are negative, so `(sin 3 alpha)/(cos 2 alpha)` is positive, therefore A`rarr r. |
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| 16. |
If `sin^ 4 x/2+cos^4 x/3 =1/5` thenA. `tan^(2) x = (2)/(3)`B. `(sin^(8)x)/(8) + (cos^(8)x)/(27) = (1)/(125)`C. `tan^(2)x=(1)/(3)`D. `(sin^(8)x)/(8)+(cos^(8)x)/(27)=(2)/(125)` |
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Answer» Correct Answer - A::B `(sin^(4)x)/(2)+(cos^(4)x)/(3)=(1)/(5)implies(sin^(4)x)/(2)+((1-sin^(2)x)^(2))/(3)=(1)/(5)` `implies(sin^(4)x)/(2)+(1+sin^(4)x-2sin^(2)x)/(3)=(1)/(5)` `implies 5 sin^(4)x-4sin^(2)x+2=(6)/(5)` `implies 25sin^(4)x-20sin^(2)x+4=0` `implies(5sin^(2)x-2)^(2)=0` `implies sin^(2)x = (2)/(5) cos^(2)x = (3)/(5), tan^(2)x=(2)/(3)` `therefore" "(sin^(8)x)/(8)+(cos^(8)x)/(27)=(1)/(125)` |
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| 17. |
`sqrt(3)cossec2 0^0-sec2 0^0`A. 2B. 2 sin `20^(@)//sin 40^(@)`C. 4D. `4 sin 20^(@)//sin40^(@)` |
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Answer» Correct Answer - C Given expression = `sqrt(3) cosec 20^(@)-sec 20^(@) = tan 60^(@) cosec 20^(@) - sec 20^(@)` `=(sin 60^(@) cos 20^(@) - cos60^(@).sin20^(@))/(cos60^(@).sin20^(@)cos20^(@))` `=(sin(60^(@)-20^(@)))/(cos60^(@).sin20^(@).cos20^(@)) = (sin 40^(@))/((1)/(2).sin20^(@) cos20^(@))` `=(2sin20^(@) cos 20^(@))/((1)/(2)sin 20^(@) cos 20^(@)) = 4` |
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| 18. |
For a positive integer n , `f_n(theta)=(tantheta/2)(1+sectheta)(1+sec2theta)(1+sec4theta)....(1+sec2^ntheta.)`, thenA. `f_(2) ((pi)/(16))=1`B. `f_(3) ((pi)/(32))=1`C. `f_(4) ((pi)/(64))=1`D. `f_(5) ((pi)/(128))=1` |
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Answer» Correct Answer - A::B::C::D NOTE Multiplicative loop is very important approach in IIT Mathematics `(tan.(theta)/(2))(1+sectheta)=(sintheta//2)/(sintheta//2)[1+(1)/(costheta)]` `=((sintheta//2)2cos^(2)theta//2)/((costheta//2)costheta)` `=((2sintheta//2)cos theta//2)/(costheta)=(sintheta)/(costheta)=tan theta` `therefore" " f_(n)(theta) = (tantheta//2)(1+sectheta)` `(1+sec 2 theta)(1+sec2^(2)theta)...(1+sec2^(n)theta)` `=(tan theta)(1+sec 2 theta)(1+sec2^(2) theta)...(1+sec2^(n)theta)` `=tan2 theta.(1 + sec2^(2)theta)...(1+sec2^(n)theta)` `=tan (2^(n)theta)` Now, `f_(2)((pi)/(16))=tan(2^(2).(pi)/(16))=tan((pi)/(4))=1` Therefore, (a) is the answer. `f_(3)((pi)/(32))=tan(2^(3).(pi)/(32))=tan((pi)/(4))=1` Therefore, (b) is the answer. `f_(4)((pi)/(64))=tan(2^(4).(pi)/(64))=tan((pi)/(4))=1` Therefore, (c) is the answer. `f_(5)((pi)/(128))=tan(2^(5).(pi)/(128))=tan((pi)/(4))=1` Therefore, (d) is the answer. |
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| 19. |
If `theta` and `phi` are acute angles satisfying `sin theta=1/2, cos phi=1/3` then `theta+phi=`A. `(pi/(3), (pi)/(6)]`B. `((pi)/(2), (2pi)/(3))`C. `((2pi)/(3),(5pi)/(6))`D. `((5pi)/(6),pi]` |
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Answer» Correct Answer - B Since, `sin theta = (1)/(2)` and cos `phi = (1)/(3) implies thet = (pi)/(6)` and `0 lt (cos phi - (1)/(3)) lt (1)/(2) " "["as " 0 lt (1)/(3) lt (1)/(2)]` `implies theta = (pi)/(6) and cos^(-1)(0) gt phi gt cos^(-1) ((1)/(2))` [the sign changed as cos x is decreasing between `(0, (pi)/(2))`] `implies theta = (pi)/(6) and (pi)/(3) lt phi lt (pi)/(2) implies (pi)/(2) lt theta + phi lt (2pi)/(3)` `therefore" " theta in ((pi)/(2), (2pi)/(3)) ` |
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| 20. |
If `alpha+beta=pi/2` and `beta+gamma=alpha` then `tanalpha` equalsA. `2(tan beta + tan gamma)`B. `tan beta + tangamma`C. `tanbeta + 2tan gamma`D. `2tan beta + tan gamma` |
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Answer» Correct Answer - c Given, `alpha+beta=pi//2` `implies alpha=(pi//2)-beta` `implies tan alpha = tan (pi//2-beta)` `implies tan alpha = cot beta` `implies tan alpha tan beta = 1` Again, `beta + gamma = alpha " [given]"` `implies gamma = (alpha - beta)` `implies tan gamma = tan(alpha - beta)` `implies tan gamma = (tanalpha-tanbeta)/(1+tanalphatanbeta)` `implies tan gamma = (tan alpha-tanbeta)/(1+1)` `therefore 2 tan gamma = tan alpha - tan beta` `implies tan alpha=tanbeta + 2 tan gamma` |
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| 21. |
The value of `3(cos theta-sin theta)^(4)+6(sin theta+cos theta)^(2)+4 sin^(6) theta` is where `theta in ((pi)/(4),(pi)/(2))` (a) `13-4cos^(4) theta` (b) `13-4cos^(6) theta` (c) `13-4cos^(6) theta+ 2 sin^(4) theta cos^(2) theta` (d) `13-4cos^(4) theta+ 2 sin^(4) theta cos^(2) theta`A. `13-4cos^(4)theta + 2sin^(2) theta cos^(2) theta`B. `13-4cos^(2)theta + 6 cos^(4) theta`C. `13-4cos^(2) theta + 6sin^(2) theta cos^(2) theta`D. `13-4cos^(6)theta` |
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Answer» Correct Answer - D Given expression `=3(sin theta - cos theta)^(4) + 6 (sin theta + cos theta)^(2) + 4sin^(6) theta` `=3((sin theta - cos theta)^(2))^(2)+6(sin theta + cos theta)^(2)+4(sin^(2)theta)^(2)` `=3 (1-sin 2 theta)^(2) + 6(1 + sin 2 theta) + 4 (1- cos^(2) theta)^(3) " "[because 1 + sin 2 theta = (cos theta + sin theta)^(2) and 1 - sin 2 theta = (cos theta - sin theta)^(2)]` `=3(1^(2)+sin^(2) 2 theta - 2 sin 2 theta) + 6(1 + sin 2 theta)+4(1-cos^(6) theta - 3 cos^(2) theta + 3 cos^(4) theta) " "[because (a-b)^(2) = a^(2)+b^(2)-2ab and (a-b)^(3) = a^(3)-b^(3)-3a^(2)b+3ab^(2)]` `=3 + 3 sin^(2) 2 theta - 6 sin 2 theta + 6 + 6 sin 2 theta + 4-4cos^(6) theta - 12 cos^(2) theta + 12 cos^(4) theta` `=13+3sin^(2)2theta - 4 cos^(6) theta - 12 cos^(2) theta+ 12cos^(4) theta` `=13+3(2sin theta cos theta)^(2) - 4 cos^(6) theta - 12 cos^(12) theta(1-cos^(2) theta)` `=13 + 12 sin^(2)theta cos^(2)theta - 4cos^(6) theta - 12 cos^(2) theta sin^(2) theta` `=13 - 4 cos^(6) theta` |
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| 22. |
Prove that : `(tanA)/(1-cotA)+(cotA)/(1-tanA)=1+secA" cosec"A`A. sin A + cos A + 1B. sec A cosec A + 1C. tan A + cot AD. sec A + cosec A |
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Answer» Correct Answer - B Given expression is `(tan A)/(1-cotA)+(cotA)/(1-tanA)` `=(sinA)/(cosA) xx (sinA)/(sinA - cosA) + (cosA)/(sinA) xx (cosA)/(cosA - sinA)` `=(1)/(sinA - cosA){(sin^(3)A-cos^(3)A)/(cosA sinA)}` `=(sin^(2)A+sinA cosA + cos^(2)A)/(sinA cosA)` `=(1+sin A cos A)/(sin A cos A) = 1 + sec A " cosec " A` |
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| 23. |
Let `f_k(x) = 1/k(sin^k x + cos^k x)` where `x in RR` and `k gt= 1.` Then `f_4(x) - f_6(x)` equalsA. `(1)/(12)`B. `(5)/(12)`C. `(-1)/(12)`D. `(1)/(4)` |
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Answer» Correct Answer - A We have, `f_(k)(x) = (1)/(k)(sin^(h)x+ cos^(k)x), k = 1, 2, 3, ...` `therefore f_(4)(x) = (1)/(4)(sin^(4)x + cos^(4) x)` `=(1)/(4)((sin^(2)x + cos^(2)x)^(2)-2sin^(2)xcos^(2)x)` `=(1)/(4)(1-(1)/(2)(sin2x)^(2))=(1)/(4)-(1)/(8)sin^(2)2x` and `f_(6)(x)=(1)/(6)(sin^(6)x + cos^(6)x)` `=(1)/(6){(sin^(2)x+cos^(2)x)^(3)-3sin^(2) x cos^(2)x (sin^(2)x + cos^(2)x)}` `=(1)/(6){1-(3)/(4)(2sin x cos x)^(2)}=(1)/(6)-(1)/(8)sin^(2)2x` Now, `f_(4)(x) - f_(6)(x)=(1)/(4)-(1)/(6)=(3-2)/(12)=(1)/(12)` |
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