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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.
| 101. |
`( cos 4x + cos 3x + cos 2x )/( sin 4x + sin 3x + sin 2x) = cot3x` |
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Answer» LHS `=(cos4x+cos3x+cos2x)/(sin4x+sin3x+sin2x)` `=(cos3x+(cos4x+cos2x))/(sin3x+(sin4x+sin2x))` `=(cos3x+2cos(4x+2x)/(2)cos(4x-2x)/(2))/(sin3x+2sin(4x+2x)/(2)cos(4x-2x)/(2))` `(cos3x+2cos3xcosx)/(sin3x+2sin3xcosx) = (cos3x(1+2cosx))/(sin3x(1+2cosx))` `(cos3x)/(sin3x) = cot3x` = RHS. Hence Proved. |
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| 102. |
`tanx=sqrt(3)` |
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Answer» `therefore tanx=sqrt(3)=tan60^(@)=tanpi/3` `therefore x=pi/3` Again, `tanx=sqrt(3)=tan(pi+pi/3)=tan(4pi)/(3)` `x=(4pi)/(3)` Therefore, for the equation `tanx=sqrt(3)`, principal solution is `x=pi/3` or `(4pi)/(3)`. and general solutions is `x=npi+pi/3, n in Z` Ans. |
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| 103. |
`cot4x(sin5x+sin3x)= cotx(sin5x-sin3x)` |
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Answer» LHS `=cot4x(sin5x+sin3x)` `=(cos4x)/(sin4x)[2.sin(5x+3x)/(2) cos(5x-3x)/(2)]` `=(cos4x)/(sin4x).2sin4x.cosx.=2cos4x.cosx` `therefore cot4x(sin5x+sin3x)=2cos4xcosx`………………..(1) `=(cosx)/(sinx)(2.cos(5x+3x)/(2)sin(5x-3x)(2))` `=(cosx)/(sinx). 2cos4x.sinx=2cos4xcosx` `therefore cotx(sinx-sin3x)=2cos4xcosx`............(2) From equations, (1) and (2), `cot4x(sin5x+sin3x)=cotx(sin5x-sin3x)` LHS =RHS Hence proved. |
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| 104. |
Prove that: `s in" "x" "+" "s in" "3x" "+" "s in" "5x" "+" "s in" "7x" "=" "4" "cos" "x" "cos" "2x" "s in" "4x` |
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Answer» LHS `=sinx+sin3x+sin5x+sin7x = (sin7x+sin5x)+(sin5x+sin3x)` `=2sin(7x+x)/(2)cos(7x-x)/(2) + 2sin(5x+3x)/(2)cos(5x-3x)/(2)` `=2sin4xcos3x+2sin4xcosx` `=2sin4x(cos3x+cosx)` `=2sin4x[2cos(3x+x)(2)cos(3x-x)/(2)]` `=2sin4x[2cos2xcosx]` `=4cosxcos2xsin4x`= RHS Hence Proved. |
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| 105. |
Solve the equation `4cos^(2)theta+sqrt(3)=2(sqrt(3)+1)costheta`. |
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Answer» `4cos^(2)theta+sqrt(3)=2(sqrt(3)+1)costheta` `rArr 4cos^(2)theta+sqrt(3)=2sqrt(3)costheta+2costheta` `rArr 4cos^(2)theta-2costheta-2sqrt(3)costheta-2sqrt(3)costheta+sqrt(3)=0` `rArr 2costheta(2costheta-1)-sqrt(3)(2costheta-1)=0` `rArr (2costheta-1)(2costheta-sqrt(3))=0` Now `2costheta-1=0` `rArr costheta=1/2=cospi/3` `rArr theta=2npi+-pi/3`. Ans. and `2costheta-sqrt(3)=0` `rArr costheta=sqrt(3)/2=cospi/6` `rArr theta=2npi+-pi/6`. |
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| 106. |
`sin2x+cosx=0` |
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Answer» Given equation: `sin2x+cosx=0`, or `2sinxcosx+cosx=0`, or `cosx(2sinx+1)=0` `rArr cosx=0` or `2sinx+1=0` If `cosx=0` then `x=(2n+1)pi/2`, where `n in z`, and `2sinx+1=0, rArr sinx=-1/2=sin(pi+pi/6)` `rArr sinx=sin(pi + pi/6)` `rArr sinx=sin(7pi)/(6)` `rArr x=npi+(-1)^(n)(7pi)/(6)`, where `n in Z` Therefore, the general solution of given equation is `x=npi+(-1)^(n)(7pi)/(6)` or `(2n+1)pi/2, n in Z` Ans. |
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| 107. |
`(tan(pi/4+x))/(tan(pi/4-x)) = ((1+tanx)/(1-tanx))^(2)` |
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Answer» LHS, `therefore tan(pi/4+x) = (tanpi/4+tanx)/(1-tanpi/4.tanx)` `therefore tan(pi/4+x) = (1+tanx)/(1-tanx)`………….(1) Similarly, `tan(pi/4-x) = (1-tanx)/(1+tanx)`………..(2) Divide equation (1) by equation (2), `therefore (tan(pi/4+x))/(tan(pi/4-x)) = ((1+tanx)/(1-tanx))/((1-tanx)/(1+tanx))` `=(1+tanx)/(1-tanx) xx (1+tanx)/(1-tanx)` `=(1+tanx)/(1-tanx)^(2)`= RHS Hence Proved. |
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| 108. |
`sin2x+2sin4x+sin6x= 4cos^(2)xsin4x` |
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Answer» LHS `=sin2x+2sin4x+sin6x` `=sin4x+[2sin(6x+2x)/(2)cos(6x-2x)/(2)]` `=2sin4x+2sin4xcos2x=2sin4x[1+cos2x]` `=2sin4x[1+2cos^(2)x-1]=2sin4x.2cos^(2)x` `=4cos^(2)x.sin4x`=RHS Hence Proved. |
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| 109. |
`sin^(2)6x-sin^(2)4x=sin2xsin10x` |
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Answer» LHS `=sin^(2)6x-sin^(2)4x` `=1/2[2sin^(2)6x-2sin^(2)4x]` `=1/2[(1-cos12x)=(1-cos8x)]` `[therefore 2sin^(2)nx=(1-cos2nx)]` `=1/2[cos8x-cos12x]` `1/2[2sin(8x+12x)/(2) sin(12x-8x)/(2)]` `=sin10xsin2x=sin2xsin10x` =R.H.S. Hence Proved. |
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| 110. |
`(cos(pi-x)cos(-x))/(sin(pi-x)cos(pi/2+x))=cot^(2)x` |
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Answer» LHS. `=(cos(pi+x).cos(-x))/(sin(pi-x).cos(pi/2+x)) = (-cos x.cos x)/(sinx.(-sinx))` `=(-cos^(2)x)/(-sin^(2)x)=cot^(2)x`= RHS Hence Proved. |
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| 111. |
Prove that:`cos^2 2x-cos^2 6x=sin4xsin8x` |
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Answer» LHS `=cos^(2)2x-cos^(2)6x` `=1/2[2cos^(2)2x-2cos^(2)6x]` `=1/2[cos4x-cos12x]` `=1/2[2sin(4x+12x)/(2)sin(12x-4x)/(2)]` `=sin8xsin4x=sin4xsin8x` = R.H.S. Hence Proved. |
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| 112. |
`cos((3pi)/(2) +x) cos(2pi+x)[cot(3pi)/(2)-x+cot(2pi+x)]=1` |
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Answer» LHS `=cos((3pi)/(2)+x)cos(2pi+x)[(cos(3pi)/(2)-x)+cot(2pi+x)]` `=sinx.cosx[tanx+cotx]` `sinx.cosx[(sinx)/(cosx)+(cosx)/(sinx)]` `=sin^(2)x+cos^(2)x=1`= R.H.S Hence Proved. |
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| 113. |
`cos((3pi)/(4)+x)-cos((3pi)/(4)-x) = -sqrt(2)sinx` |
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Answer» LHS `=cos((3pi)/(4)+x)-cos((3pi)/(4)-x)` `=1/2[2sin^(2)6x-2sin^(2)4x]` `[therefore 2sin^(2)nx=(1-cos2nx)]` `=1/2[cos8x-cos12x]` `=1/2[cos8x-cos12x)]` `1/2[2sin(8x+12x)/(2)sin(12x-8x)/(2)]` `=sin10xsin2x=sin2xsin10x`= RHS |
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| 114. |
`sin(n+1)xsin(n+2)x+cos(n+1)xcos(n+2)x=cosx` |
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Answer» LHS `=cos(n+1)xcos(n+2)x+sin(n+1)xsin(n+2)x` `=cos{(n+1)x-(n+2)x}` `=cos(-x)=cosx` = R.H.S. Hence Proved. |
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| 115. |
`cos3thetacos5theta-cos7thetacos9theta=0` |
| Answer» `theta=(npi)/(12), theta=(npi)/(4)` | |
| 116. |
Find the area of `DeltaABC`, if a=10, `b=7sqrt(2)`m and `sqrt( c)=45^(@)` |
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Answer» `Delta=1/2absinC` `=1/2 xx 10 xx 7sqrt(2) xx sin45^(@)` `=5 xx 7sqrt(2) xx 1/sqrt(2)` =35 sq. metre. Ans. |
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| 117. |
Prove that: `sin(45^(@)-A)=1/2cos2A` |
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Answer» LHS `=sin(45^(@)+A).sin(45^(@)-A)` `=1/2[2sin(45^(@)+A).sin(45^(@)-A)]` `=1/2[cos(45^(@)+A)-(45^(@)-A)-cos(45^(@)+A)+(45^(@)-A)]` `=1/2[cos2A-cos90^(@)]` `=1/2(cos2A=0)` `=1/2cos2A`= RHS Hence proved. |
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| 118. |
If a=5, b=12 and c=13, find tanA. |
| Answer» Correct Answer - `5/12` | |
| 119. |
Prove that: `cotAcot2A-cot2Acot3A-cot3AcotA=1` |
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Answer» `3A=2A+A` `rArr cot3A=cot(2A+A)` `rArr cot3A=(cot2AcotA-1)/(cotA+cot2A)` `rArr cot3A cotA+cot2Acot3A=cotAcot2A-1` `rArr 1= cotAcot2A-cot2Acot3A-cot3AcotA` Hence Proved. |
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| 120. |
Prove that: `cot2A + tanA= cotA - cot 2A` |
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Answer» LHS `=cot2A+tanA` `=(cos2A)(sin2A)+(sinA)/(cosA)` `=(cos2AcosA+sin2AsinA)/(sin2AcosA)` `(cos(2A-A))/(sin2AcosA)=(sin(2A-A))/(sin2AsinA)` `=(sin2AcosA-cos2AsinA)/(sin2AsinA)` `=(sin2AcosA)/(sin2AsinA)-(cos2AsinA)/(sin2AsinA)` `=cotA-cot2A`= RHS Hence Proved. |
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| 121. |
If `tanA=m/(m-1)` and `tan(B)=n`, find the values of `tan 2A` and `tan 2B`. |
| Answer» `tan2A=(m+n)/(1-mn)`, `tan2B=(m-n)/(1+mn)` | |
| 122. |
In `DeltaABC, a=17,b=8,c=15`, find `sinB/2`. |
| Answer» Correct Answer - `1/sqrt(2)` | |
| 123. |
Solve: `tan2thetacottheta=1` |
| Answer» Correct Answer - `theta=npi` | |
| 124. |
In `DeltaABC`, a =125, b=62 and c=123, find the value of sinB. |
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Answer» a=125, b=62, c=123 `therefore 2s=a+b+c` `=125+62+123` `=310` `rArr s=155` Now `sinB/2=sqrt((s-a)(s-c))/(ac)` `=sqrt((155-125)(155-123))/(125 xx 123)` `=sqrt(30 xx 32)/(123 xx 123)` and `cosB/2=sqrt(s(s-b))/(ac)= sqrt(155 xx (155-62))/(125 xx 123)` `=sqrt((155 xx 93)/(125 xx 123))` and `cosB/2=sqrt((s(s-b))/(ac)) = sqrt((155 xx(155-62))/(125 xx 123)` `=sqrt((155 xx 93))/(125 xx 123)` `therefore sinB=2sinB/2 cosB/2` `=2sqrt((30 xx 32)/(125 xx 123) xx (155 xx 93)/(125 xx 123))` `(2 xx 3720)/(125 xx 23) = 496/1025` |
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| 125. |
Solve: `cot2theta=tantheta` |
| Answer» `theta=1/3(npi+pi2)` | |
| 126. |
In `DeltaABC`, prove that: `cot A/2+cot B/2+cot C/2=((a+b+c)^(2))/(4Delta)` |
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Answer» LHS `=cotA/2+cotB/2+cotC/2` `=sqrt((s(s-a))/((s-b)(s-c))+sqrt((s(s-b))/((s-a)(s-c)) + sqrt((s(s-c))/((s-a)(s-b))` `=(sqrt(s)[(s-a)+(s-b)+(s-c)])/(sqrt((s-a)(s-b)(s-c)))` `(sqrt(s)[(s-a)+(s-b)+(s-c)])/(sqrt((s-a)(s-b)(s-c)))` `(sqrt(s).sqrt(s)[3s-(a+b+c)])/(sqrt(s)(s-a)(s-b)(s-c))` `(s.(3s-2s))/(Delta)=s^(2)/Delta=(4s^(2))/(4Delta)` `=(2s)^(2)/(4Delta)=(a+b+c)^(2)/(4Delta)`= RHS Hence Proved. |
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| 127. |
If `secA=-2` and A lies in third quadrant, find the value of `(4 cot^(2)A-3sin^(2)A)`. |
| Answer» Correct Answer - `-11/12` | |
| 128. |
In `DeltaABC`, prove that: `asinA-bsinB=csin(A-B` |
| Answer» Correct Answer - `7/25` | |
| 129. |
`sintheta+sin7theta=sin4theta` |
| Answer» `theta=(npi).(4), theta=1/3(2npi+-pi/3)` | |
| 130. |
In `DeltaABC`, a=16, b=12 and `angleB=30^(@)`, find sinA. |
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Answer» From sine rule `a/(sinA)=b/(sinB)` `rArr sinA=(asinB)/(b) = (16 sin30^(@))/(12)` `=(16 xx 1)/(12 xx 2)=2/3`. Ans. |
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| 131. |
In `DeltaABC`, a=4, b=6 and `angleB=30^(@)`, evaluate sinA. |
| Answer» Correct Answer - `1/3` | |
| 132. |
`cosec (-1410^@)` |
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Answer» `cosec "(-1410^(@))` `=-"cosec"1410^(@)` `=-"cosec"(3 xx 360^(@) +330^(@))` `=-"cosec"330^(@)` `="cosec"(360^(@)-30^(@))` `=-"cosec"30^(@)` =2 Ans. |
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| 133. |
If `sinA=-3/5` and A lies in third quadrant, find the remaining trignometric ratios. |
| Answer» `cosA=-4/5, tanA=3/4, "cosec"A=-5/3, secA=-5/4, cotA=4/3` | |
| 134. |
Evaluate: `"cosec"(-1410)^(@)` |
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Answer» `=-"cosec"1410^(@)` `=-"cosec"(3 xx 360^(@) + 330^(@))` `=-"cosec"330^(@)` `=-"cosec"(360^(@)-30^(@))` `="cosec"30^(@)=2` Ans. |
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| 135. |
Find the values of the following : i) `tan (19pi)/3` ii) `sec(-22pi)/3` |
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Answer» i) `tan(19pi)/3= tan(19 xx 180^(@))/(3)` `=tan(1140^(@))` `=tan(3 xx 360^(@) +60^(@))` `=tan60^(@)=sqrt(3)` ii) `sec(-22pi)/(3)=sec(22pi)/(3)=sec(22 xx 180^(@))/(3)` `=sec1320^(@)` `=sec(3 xx 360^(@) + 240^(@))` `=sec240^(@)=sec(180^(@)+60^(@))` `=-sec60^(@)=-2` Ans. |
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| 136. |
`cosx=-1/3`, x in quadrant III. Find the values of other five trignometric functions |
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Answer» `therefore cosx=-1/3` `therefore 2cos^(2)x/2-1=-1/3` or `2cos^(2)x/2=1-1/3=2/3`, or `cos^(2)x/2=1/3` or `cosx/2=+-1/sqrt(3)=+-sqrt(3)/3` Again, `cosx=-1/3` `rArr 1-2sin^(2)x/2=-1/3` or `2sin^(2)x/2=1+1/3=4/3` or `sin^(2)x/2=2/3`, or `sinx/2=+sqrt(2)/(3)`, or `sinx/2=+-sqrt(6)/3` `therefore` x lies in III quadrant. `rArr` The value of x will be lie between `pi` and `(3pi)/(2)` i.e., `pi lt x lt (3pi)/(2)` Then the value of `x/2` will lie between `pi/2` and `(3pi)/(4)` i.e., `pi/2 lt x/2 lt (3pi)/(4)` Then `x/2`, lies in 2nd quadrant in which `sinx/2` is positive while `cosx/2` and `tanx/2` are negative. `therefore sinx/2=sqrt(6)/(3)`, then `cosx/2=-sqrt(3)/3` and `tanx/2=(sqrt(6)//3)/(-sqrt(3)//3)=-sqrt(2)` Therefore, `sinx/2=sqrt(6)/(3), cosx/2=-sqrt(3)/3` and `tanx/2=-sqrt(2)` |
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| 137. |
If `sinA+cosA=0` and A lies in 4th quadrant, find the values of `sinA` and `cosA.` |
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Answer» `sinA+cosA=0` `rArr sinA=-cosA` `rArr (sinA)/(cosA)=-1` `rArr sec^(2)A=1+tan^(2)A` `=1+(-1)^(2)=2` secA`=sqrt(2)`, (sec is positive in 4th quadrant) `rArr cosA=1/(secA)=1/sqrt(12)` Ans. `rArr cosA=1/(secA)=1/sqrt(2)`. and `sinA=(sinA)/(cosA). cosA=tanA.cosA` `=(-1).1/sqrt(2)=-1/sqrt(2)` Ans. and sinA`=(sinA)/(cosA).cosA=tanA. cosA` |
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| 138. |
If `"cosec"A=5/4` and A lies in first quadrant, find the remaining trignometric ratios. |
| Answer» `sinA=4/5, cosA=3/5, tanA=4/3, secA=5/3, cotA=3/4` | |
| 139. |
Find `sin``x/2``,cos``x/2`and `tan``x/2`of the following : `tanx=-4/3,x`in quadrant II |
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Answer» `therefore tanx=-4/3` `therefore (2tanx/2)/(1-tan^(2)x/2)=-4/3` or `-4+4tan^(2)x/2=6tanx/2` or `2tan^(2)x/2-3tanx/2-2=0` or `2tanx/2(tanx/2-2)+1(tanx/2-2)=0` then `(2tanx/2+1=0)` or `(tanx/2-2)=0` If `2tanx/2+1=0`, then `tanx/2=-1/2`, and if `tanx/2-1=0`, then `tanx/2=2` But x lies in second quadrant from which sin x positive, so the value `-1/2` of `tanx/2` is not acceptable. Then `tanx/2=2` So, `cosx=(1-tan^(2)x/2)/(1+tan^(2)x/2) = (1-(2)^(2))/(1+(2)^(2))=-3/5` `therefore cosx=-3/5` `rArr 1-2sin^(2)x/2 = -3/5` or `2cos^(2)x/2-1=-3/5` If `1-2sin^(2)x/2=-3/5` `2sin^(2)x/2=1+3/5=8/5` or `sin^(2)x/2=4/5` or `sinx/2=2/sqrt(5)` If `2cos^(2)x/2-1=-3/5` then `2cos^(2)x/2-1=-3/5` then `2cos^(2)x/2=1-3/5=2/5` or `cos^(2)x/2=1/5` or `cosx/2=1/sqrt(5)` Therefore, `sinx/2=2/sqrt(5), cosx/2=1/5` and `tanx/2=2` Ans. |
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| 140. |
Find the value of other five trigonometric function `tanx=-5/(12)`, x lies in second quadrant. |
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Answer» Given, that `tanx=-5/12`, and x lies in second quadrant. `therefore tanx=-5/12 rArr cotx=1/(tanx)` `rArr cotx=1/(-5/12) = 1 xx -12/5 = -12/5` `(therefore` x lies in second quadrant, therefore, sin and "cosec" ratio willl be positive and the remaining ratio tan, cot and sec will be negative.) then `sec^(2)x=-13/12` `therefore cos=1/(secx)` `rArr cosx=1/(-13/12) = 1 xx (-12/13)` `rArr cosx=-12/13` `therefore "cosec"^(2)x=cot^(2)x` `therefore "cosec"^(2)x=1+(-12/5)^(2)= 1+144/25=169/25=(13/5)^(2)` `rArr "cosec"x=13/5` `therefore sinx=1/("cosec"x) rArr sinx=1/(13/5) = 1 xx 5/13` `therefore sinx=5/13` Therefore, `sinx=5/13`, `"cosec"x=13/5, cosx=-12/13` `cotx=-12/5, secx=-13/12`. Ans. |
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| 141. |
Find the value of other five trigonometric function `sinx=3/5`, x lies in second quadrant. |
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Answer» Correct Answer - c Given that `sinx=3/5` and x lies in second quadrant. `sinx=3/5` `therefore "cosec"x = 1/(sinx) = 1/(3//5)=1 xx 5/3=5/3` `therefore "cosec"x=5/3`. `(therefore` x lies in second quadrant, so sin ratio will be positive and the remaining ratio tan, cot, cos and sec will be negative). `therefore sin^(2)x+cos^(2)=1 rArr (3/5)^(2) + cos^(2)x=1` `rArr 9/25 + cos^(2)x=1` `cos^(2)x=1-9/25 rArr cos^(2)x=16/25` `rArr cosx =+-4/5` `therefore secx=1/(cosx)` `rArr tanx=(sinx)/(cosx) = (3//5)/(-4//5) = 3/5 xx (-5/4) = -3/4` `therefore cotx=(cosx)/(sinx)` `therefore cotx = (-4//5)/(3//5) = -4/5 xx 5/3 =-4/3` Therefore, `"cosec"x=5/3, cosx=-4/5`, `secx=-5/4, tanx=-3/4` and `cotx=-4/3` Ans. |
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| 142. |
Find the value of other five trigonometric function `secx=(13)/5`, x lies in fourth quadrant. |
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Answer» Given that `secx=13/5` and x lies in fourth quadrant. `therefore secx=13/5` and x lies n fourth quadrant. `therefore secx=13/5` and `cosx=1/(secx) rArr cosx=5/13` `(therefore` x lies in fourth quadrant, therefore sec and cos ratio will be positive and remaining all ratio will be negative.) `therefore sin^(2)x+cos^(2)=1` `therefore sin^(2)x+(5/13)^(2)=1 rArr sin^(2)x=1-25/169 = 144/169` `rArr sinx=-12/13` `therefore "cosec"x=-1/(sinx)` `rArr "cosec"x=1/(-12//13) = 1 xx (-13/12) = -13/12` `therefore tanx=(sinx)/(cosx) = (-12//13)/(5//13) = -12/5` `therefore cotx=1/(tanx) rArr cotx=1/(-12//5)` `cotx=-5/12` Ans. Therefore, `cosx=5/13, sinx=-12/13, "cosec "x=-13/12`, `tanx=-12/5, cotx=-5/12` Ans. |
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| 143. |
In a triangle `A B C ,/_C=60^0,`then prove that: `1/(a+c)+1/(b+c)=3/(a+b+c)dot` |
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Answer» Let `1/(a+b)+1/(b+c)=3/(a+b+c)` `rArr (b+c+a+b)/((a+b)(b+c))=3/(a+b+c)` `rArr (a+2b+c)(a+b+c)=3(a+b)(b+c)` `rArr a^(2)+2b^(2)+c^(2)+3ab+3bc+2ac = 3ab + 3b^(2)+3ac+3bc` `rArr a^(2)+c^(2)-b^(2)=ac` `rArr (a^(2)+c^(2)-b^(2))/(2ac)=1/2` `rArr cosB=cos60^(@)` `rArr B=60^(2)` `therefore` In `DeltaABC, angleB=60^(@)` `rArr 1/(a+b)+1/(b+c)=3/(a+b+c)`. Hence Proved. |
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| 144. |
Convert `60^(@)` angle into radian. |
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Answer» `60^(@) = 60 xx pi/180` radian `=pi/3` radian. |
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| 145. |
Convert 11 radian into degree. |
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Answer» 11 radian `=(11 xx 180^(@)/pi)^(@)` `=(11 xx (180 xx 7)/(22))^(@)` `=630^(@)` |
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| 146. |
Prove that: `cos^(2)48^(@)-sin^(2)12^(@)=(sqrt(5)+1)/(8)` |
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Answer» LHS `=cos^(2)48^(@)-sin^(2)12^(@)` `=cos(48^(@)+12^(@))cos(48^(@)-12^(@))` `=cos60^(@)cos36^(@)` `=1/2 xx (sqrt(5)+1)/(8)=`RHS Hence Proved. |
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| 147. |
Prove that: `sin12^(@)sin48^(@)sin54^(@)=1/8` |
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Answer» LHS `=sin12^(@)sin48^(@)sin54^(@)` `=1/2.(2sin48^(@)sin12^(@)).sin(90^(@)-36^(@))` `=1/2[cos(48^(@)-12^(@))-cos(48^(@)+12^(@))].cos36^(@)` `=1/2[cos(48^(@)-12^(@))-cos(48^(@)+12^(@))].cos36^(@)` `=1/2[cos36^(@)-cos60^(@)].cos36^(@)` `=1/2[(sqrt(5)+1)/(4)-1/2].((sqrt(5)+1)/(3))` `=1/2(sqrt(5)+1-2)/(4).(sqrt(5)+1)/(4)` `=((sqrt(5)-1)(sqrt(5)+1))/(32)` `=(5-1)/(32)=4/(32)=1/8` = RHS Hence Proved. |
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| 148. |
Prove that: `sin6^(@)sin42^(@)sin66^(@)sin78^(@)=1/16` |
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Answer» LHS `=sin6^(@)sin42^(@)sin66^(@)sin78^(@)` `=1/4(2sin66^(@)sin6^(@)).(2sin78^(@)sin42^(@))` `=1/4[cos(66^(@)-6^(@))-cos(66^(@)+6^(@))] [cos(78^(@)-42^(@))-cos(78^(@)+42^(@))]` `=1/4[cos60^(@)-cos72^(@)][cos36^(@)-cos120^(@)]` `=1/4[1/2-cos(90^(@)-18^(@))][cos36^(@)-cos(90^(@)+30^(@))]` `=1/4[1/2-sin18^(@)][cos36^(@)+sin30^(@)]` `=1/4[1/2-(sqrt(5)-1)/(4)][(sqrt(5)+1)/(4)+1/2]` `=1/4[(2-sqrt(5)+1)/(4)][(sqrt(5)+1+2)/(4)]` `=3-sqrt(5)(3+sqrt(5))/(64)=(9-5)/(64)` `=4/64=1/16`= RHS Hence Proved. |
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| 149. |
In `DeltaABC`, prove that: `sin2A+sin2B+sin2C=4sinAsinBsinC` |
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Answer» LHS=`sin2A+sin2B+sin2C` `=2sin(180^(@)-C)cos(A-B)+2sinC).cos{180^(@)-A(A+B)}(therefore A+B+C-180^(@))` `=2sinC.cos(A-B)-2sinC.cos(A+B)` `=2sinc[cos(A-B)-cos(A+B)]` `=2sinC.2sinAsinB` `=4sinAsinBsinC`=RHS Hence Proved. |
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| 150. |
If `tanA=xtanB`, prove that: `(x+1)sin(A-B)=(x-1)sin(A+B)` |
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Answer» `tanA= xtanB` `rArr x=(tanA)/(tanB)=((sinA)/(cosA))/((sinB)/(cosB))=(sinAcosB)/(sinBcosB)` Now `(x+1)/(x-1)=((sinAcosB)/(sinBcosA)+1)/((sinAcosB)/(sinBcosA)-1)` `=(sinAcosB+sinBcosA)/(sinAcosB-sinBcosA)` `=(sin(A+B))(sin(A-B))` `rArr (x+1)sin(A-B)=(x-1)sin(A+B)` Hence Proved. |
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