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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. |
If `cos25^0+sin25^0=p ,`then `cos50^0`is`sqrt(2-p^2)`(b) `-sqrt(2-p^2)``psqrt(2-p^2)`(d) `-psqrt(2-p^2)`A. `sqrt(2-p^(2))`B. `-sqrt(2-p^(2))`C. `p*sqrt(2-p^(2))`D. `-p*sqrt(2-p^(2))` |
| Answer» Correct Answer - C | |
| 102. |
If `acostheta+bsintheta=m and asintheta-bcostheta=n`, then show that ` a^(2)+b^(2)=m^(2)+n^(2)`. |
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Answer» Given that, `" "acostheta+bsintheta=m" "…(i)` ltBrgt and `" "asintheta-bcostheta=n" "...(ii)` On squaring and adding of Eqs. (i) and (ii), we get `" "m^(2)+n^(2)=(acostheta+bsintheta)^(2)+(asintheta-bcostheta)^(2)` `" "=a^(2)cos^(2)theta+b^(2)sin^(2) theta+2ab sintheta.costheta+a^(2)sin^(2)theta+b ^(2)cos^(2)theta-2ab""sintheta*costheta` `rArr" "m ^(2)+n^(2)=a^(2)(cos^(2)theta+sin^(2)theta) +b^(2)(sin^(2)theta+ cos^(2) theta)` `rArr" "m^(2) +n^(2)=a^(2) +b ^(2)" "` Hence proved. |
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| 103. |
If `x/acostheta+y/bsintheta=1,x/asintheta-y/bcostheta=1`, then eliminate `theta`. |
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Answer» Correct Answer - `x^2/a^2+y^2/b^2=2` Squaring and adding, we have `x^2/a^2+y^2/b^2=2` |
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| 104. |
In `triangleABC`, if `a=18, b=24, c=30`, then `tan((A)/(2))=`A. `(1)/(3)`B. `(2)/(3)`C. `(4)/(3)`D. `1` |
| Answer» Correct Answer - A | |
| 105. |
In `triangleABC`, if `a=18, b=24, c=30`, then `cos((A)/(2))=`A. `(4)/(sqrt(10))`B. `(3)/(sqrt(10))`C. `(2)/(sqrt(10))`D. `(1)/(sqrt(10))` |
| Answer» Correct Answer - B | |
| 106. |
If `x=secphi -tanphi and y="cosec" phi+cotphi`, then show that `xy+x-y+1=0.` |
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Answer» Given that, `" "x=secphi-tanphi" "...(i)` and `" "y="cosec"phi+cotphi" "...(ii)` `" "1*xy=(secphi-tanphi)("cosec"phi+cotphi )` `rArr" "xy=secphi*"cosec"phi-"cosec"phi*tanphi+secphi*cot phi -tanphi*cotphi` `rArr" "xy=secphi*"cosec"phi-(1)/(cosphi)+(1)/(sinphi)-1` `rArr" "1+xy=secphi"cosec"phi-sec phi+"cosec"phi" "...(iii)` From Eqs. (i) and (ii), we get `" "x-y=secphi-tanphi-"cosec"phi-cotphi` `rArr" "x-y=secphi-"cosec"phi-(sinphi)/(cosphi)-(cosphi)/(sinphi)` `rArr" "x-y=secphi-"cosec"phi-((sin^(2)phi+cos^(2)phi)/(sinphi*cosphi))` `rArr" "x-y=sec phi-"cosec"phi-"cosec"phi*secphi` `rArr" " x-y=-(secphi*"cosec"phi-secphi+"cosec"phi)` `rArr" "x-y=-(xy+1)" "` [from Eq. (iii)] `rArr" "xy+x-y+1=0" "` Hence proved. |
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| 107. |
In `triangleABC`, if `angleC=90^(@)` then `tanA+tanB`=A. `(a^(2))/(bc)`B. `(b^(2))/(ca)`C. `(c^(2))/(ab)`D. `a+b` |
| Answer» Correct Answer - C | |
| 108. |
If : `A+B = 90^(@),and tanA=3/4 "then" : cot B =`A. -1B. 0C. 1D. `sqrt2` |
| Answer» Correct Answer - C | |
| 109. |
In `triangleABC`, if `a=18, b=24, c=30`, then `tanA=`A. `1`B. `(3)/(4)`C. `(2)/(4)`D. `(1)/(4)` |
| Answer» Correct Answer - B | |
| 110. |
If `sin A=11//61, "where" 0ltAlt(pi//2),` `"then : (secA+tanA)/(cotA-cscA)=`A)`66//13` B)`-13//66` C)`-66//15` D)`-66//5`A. `66//13`B. `-13//66`C. `-66//15`D. `-66//5` |
| Answer» Correct Answer - D | |
| 111. |
In `triangleABC`, `(b^(2)+c^(2)-a^(2))tanA=`A. `2abc`B. `abc`C. `2bcsinA`D. `bcsinA` |
| Answer» Correct Answer - C | |
| 112. |
For acute angle `theta`, prove the following: (i) `sec^2thetacosecthetage4` (ii) `sec^2theta+cosec^2thetage4` |
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Answer» Using `A.M.geG.M.` `(sin^2theta+cos^2thet)/2ge(sin^2thetacos^2theta)^(1//2)` `rArr 1/2ge(sin^2thetacos^2theta)^(1//2)rArrsin^2thetacos^2thetale1/4` `=sec^2thetacosec^2thetage4rArr(sin^2theta+cos^2theta)/(sin^2thetacos^2theta)ge4` `rArr sec^2theta+cosec^2thetage4` |
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| 113. |
Find the general solution of the equation `5cos^(2)theta+7sin^(2)theta-6=0`. |
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Answer» Given equation, `5cos^(2)theta+7sin^(2)theta-6=0` `rArr" "5cos^(2)theta+7(1-cos^(2)theta)-6=0` `rArr" "5cos^(2)theta+7-7cos^(2)theta-6=0` `rArr" "5cos ^(2)theta+7-7cos^(2)theta-6=0" "rArr" "-2cos^(2)theta+1=0` `rArr" "2cos^(2)theta -1=0" "[because cos^(2)theta=cos^(2)alpha thereforetheta=npipmalpha]` `rArr" "cos^(2)theta=(1)/(2)` `rArr" "cos^(2)theta=cos^(2)""(pi)/(4)` `therefore" "theta=npipm(pi)/(4)` |
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| 114. |
If `theta` lies in the first quardrant and `costheta=(8)/( 17 )`, then find the value of `cos(30^(@)+theta)+cos(45^(@)-theta)+cos(120^(@)-theta)`. |
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Answer» Given that, `" "cos3theta=(8)/(17)rArrsintheta=sqrt(1-(64)/(289))` `rArr" "sintheta=sqrt((289-64)/(289))rArrsintheta=pm(15)/(17)` `rArr" "sintheta=(15)/(17)" "` [since, `theta` lies in the first quadrant] Now, `cos(30^(@)+theta)+cos(45^(@)-theta)+cos(120^(@)-theta)` `" "=cos(30^(@)+theta)+cos(45 ^(@)-theta)+cos(90^(@)+30^(@)-theta)` `" "=cos(30^(@)+theta)+cos(45^(@) -theta)-sin(30^(@)-theta)` `" "=cos30^(@)costheta-sin 30^(@) sin theta+cos45^(@) cos theta+sin45^(@) sintheta-sin30^(@) costheta+cos30^(@)sintheta` `" "=(sqrt(3) )/(2)costheta-(1)/(2) sintheta+(1)/(sqrt(2))costheta+ (1)/(sqrt (2)) sintheta-(1)/(2)costheta(sqrt(3))/(2)sintheta` `" "=((sqrt(3))/(2)+(1)/(sqrt(2))-(1)/(sqrt(2)))costheta+((1)/(sqrt(2))-(1)/(2)+(sqrt(3))/(2))sintheta` `"" =((sqrt(6)+2-sqrt(2))/(2sqrt(2)))costheta+((2-sqrt(2)+sqrt(6))/(2sqrt(2)))sintheta` `" "=((sqrt(6)+2-sqrt(2))/(2sqrt(2)))(8)/(17)+((2-sqrt(2)+sqrt(6))/(2sqrt(2)))(15)/(17)` `" "=(1)/(17(2sqrt2))(8sqrt(6)+16-8sqrt(2)+30-15sqrt(2)+15sqrt(6)` `" "=(1)/(17(2sqrt(2)))(23sqrt(6)-23sqrt(2)+46)` `" "=(23sqrt(6))/(17(2sqrt(2)))-(23sqrt(2))/(17(2sqrt(2)))+(46)/(17(2sqrt2))` `" "=(23sqrt(3))/(17(2))-(23)/(17(2))+(23)/(17sqrt(2))` `" "=(23)/(17)((sqrt(3)-1)/(2)+(1)/(sqrt(2)))` |
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| 115. |
If `tanA=(sqrt3-1)/(sqrt3+1),"then": cos A=`A. `(sqrt3-1)/(2sqrt2)`B. `(sqrt3+1)/(2sqrt2)`C. `(sqrt3+1)`D. `(sqrt3-1)` |
| Answer» Correct Answer - B | |
| 116. |
Let `f:(-1,1)vecR`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 (are)(a)`1-sqrt(3/2)`(b) `1+sqrt(3/2)`(c) `1-sqrt(2/3)`(d) `1+sqrt(2/3)` |
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Answer» It is given that, `f(cos4theta) = 2/(2-sec^2theta)` We have to find `f(1/3)`. So, we can consider, `cos4theta = 1/3` Now, `cos4theta = 2cos^2 2theta -1` `=>cos2theta = +-sqrt((1+cos4theta)/2 )` `=>cos2theta = +-sqrt((1+(1/3))/2 )` `=>cos2theta = +-sqrt(2/3)` Now, `f(1/3) = 2/(2-sec^2theta)` `=(2cos^2theta)/(2cos^2theta - 1)` `=(1+cos2theta)/(cos2theta)` `=1/(cos2theta)+1` `=+-sqrt(3/2) +1` `=>f(1/3) = 1+-sqrt(3/2)` So, option `a` and `b` are the correct options. |
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| 117. |
If : `tantheta=(1)/(7), "then" : (2sintheta+3costheta)/(3sintheta+4costheta)=`A. `(32)/(13)`B. `(23)/(13)`C. `(23)/(31)`D. `(13)/(23)` |
| Answer» Correct Answer - C | |
| 118. |
The value of `(sin300^(@).tan330^(@).sec420^(@))/(tan135^(@).sin210^(@).sec315^(@))` isA. `-1`B. 1C. `sqrt(2)`D. `sqrt(3)` |
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Answer» Correct Answer - C Given expression `=(sin(360^(@)-60^(@)).tan(360^(@)-30^(@)).sec(360^(@)+60^(@)))/(tan(180^(@)-45^(@)).sin(180^(@)+30^(@)).sec(360^(@)-45^(@)))` `=((-sin 60^(@)).(-tan 30^(@)).sec 60^(@))/((-tan 45^(@)).(-sin 30^(@)).sec 45^(@))` `=(sin 60^(@).tan30^(@).sec60^(@))/(tan 45^(@).sin 30^(@).sec 45^(@))` `= ((.sqrt(3))/(2).(1)/(sqrt(3)).2)/(1.(1)/(2).sqrt(2))=sqrt(2)` |
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| 119. |
`tan^6(pi/9)-33tan^4(pi/9)+27tan^2(pi/9)`is equal to(a)0 (b) `sqrt3` (c) 3(d)9 |
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Answer» We know, `tan 3theta = (3tantheta - tan^3theta)/(1-3tan^2theta)` If `theta = pi/9` `=>tan(pi/3) = (3tan(pi/9) - tan^3(pi/9))/(1-3tan^2(pi/9))` `=>(sqrt3(1-3tan^2(pi/9))) = (3tan(pi/9) - tan^3(pi/9))` Squaring both sides, `=>3(1+9tan^4(pi/9) - 6tan^2(pi/9)) = 9tan^2(pi/9)+tan^6pi/9 - 6tan^4(pi/9)` `=>3+33tan^4(pi/9)-27tan^2(pi/9) = tan^6(pi/9)` `=>tan^6(pi/9)-33tan^4(pi/9)+27tan^2(pi/9) = 3.` |
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| 120. |
`cos^4(pi/8)+cos^4((3pi)/8)+cos^4((5pi)/8)+cos^4((7pi)/8)=` |
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Answer» Given expression, `cos^(4)""(pi)/(8)+cos^(4)""(3pi)/(8)+cos^(4)""(5pi)/(8)+cos^(4)""(7pi)/(8)` `" "=cos^(4)""(pi)/(8)+cos^(4)""(3pi)/(8)+cos^(4)(pi-(3pi)/(8))+cos^(4)(pi-(pi)/(8))` `" "=cos^(4)""(pi)/(8)+cos^(4)""(3pi)/(8)+cos^(4)""(3pi)/(8)+cos^(4)""(pi)/(8)` `" "=2[cos^(4)""(pi)/(8)+cos^(4)""(3pi)/(8)]=2[cos^(4)""(pi)/(8)+cos^(4)""((pi)/(2)-(pi)/(8))]` `" "=2[cos^(4)""(pi)/(8)+sin^(4)""(pi)/(8)]` `" "=2[(cos^(2)""(pi)/(8)+sin^(2)""(pi)/(8))^(2)-2cos^(2)""(pi)/(8)*sin^(2)""(pi)/(8)]` `" "=2[1-2cos^(2)""(pi)/(8)*sin^(2)""(pi)/(8)] = 2-(2sin""(pi)/(8)*cos"(pi)/(8))^(2)` ` " "=2-(sin""(2pi)/(8))^(2)=2-((1)/(sqrt(2)))^(2)` `" "=2-(1)/(2)=(3)/(2)` |
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| 121. |
The expression `(tan(x-(pi)/(2)).cos((3pi)/(2)+x)-sin^(3)((7pi)/(2)-x))/(cos(x-(pi)/(2)).tan((3pi)/(2)+x))`simplifies toA. `(1+cos^(2)x)`B. `sin^(2)x`C. `-(1+cos^(2)x)`D. `cos^(2)x` |
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Answer» Correct Answer - B `(tan.(x-(pi)/(2)).cos.((3pi)/(2)+x)-sin^(3)((7pi)/(2)-x))/(cos(x-(pi)/(2)).tan((3pi)/(2)+x))` `=((-cot)(sin x)+cos^(3)x)/((sin x).(-ct x))` `=(-cos x + cos^(3)x)/(-cos x)` `=1-cos^(2)x` `=sin^(2)x` |
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| 122. |
If `tan theta=1/sqrt7`,then `(c o s e c^2theta-sec^2theta)/(c o s e c^2theta+sec^2theta)=?`A. `(1)/(2)`B. `(3)/(4)`C. `(5)/(6)`D. `(7)/(8)` |
| Answer» Correct Answer - B | |
| 123. |
The value of `cos(pi/7)+cos((2pi)/7)+cos((3pi)/7)+cos((4pi)/7)+cos((5pi)/7)+cos((6pi)/7)+cos((7pi)/7)`is1 (b) `-1`(c) 0(d) none of these |
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Answer» `=(cos(pi/7)+cos(6/7pi))+(cos(2/7pi)+cos(5/7pi))+(cos(3/7pi)+cos(4/7pi))+cospi` `=2cos(pi/2)cos(-5/14pi)+2cos(pi/2)*cos(-3/14pi)+2cos(pi/2)cos(-pi/14)+(-1)` `=0+0+0-1` `=-1`. |
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| 124. |
Find the value of `cos^2pi/(16)+cos^2(3pi)/(16)+cos^2(5pi)/(16)+cos^2(7pi)/(16)dot` |
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Answer» `=cos^2pi/16+cos^2 3/16pi+cos^2 5/16pi+cos^2 7/16pi` `=(1+cos(pi/8))/2+(1+cos(3/8pi))/2+(1+cos(5/8pi))/2+(1+cos(7/8pi))/2` `=2+cospi/8+cos3/8pi+cos(pi-3/8pi)+cos(pi-pi/8)` `=2+cospi/8+cos3/8pi-cospi/8` `=2`. |
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| 125. |
Find the value of : (i) `sin75^o` (ii) `tan15^o` |
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Answer» (i) `sin 75^@ = sin(45^@ + 30^@)` using identity `sin(a+b) = sinacosb + cosasinb` `= sin45^@cos30^@ + cos45^@sin30^@` `= 1/sqrt2 * sqrt3/2 + 1/sqrt2 * 1/2 ` `= sqrt3/(2sqrt2) + 1/(2sqrt2)` `= (sqrt3 + 1)/(2sqrt2)` (ii) `tan 15^@ = tan(45^@- 30^@)` using identity `tan(a-b) = (tana- tanb)/(1+tan a tan b)` `= (tan 45^@ - tan 30^@)/(1 + tan45^@tan30^@)` `= (1-1/sqrt3)/(1 + 1*1/sqrt3)` `= ((sqrt3 - 1)/(sqrt3))/((sqrt3 + 1)/sqrt3)` `= (sqrt3 -1)/(sqrt3+1) * (sqrt3 - 1)/(sqrt3 -1) ` `= (sqrt3 - 1)^@/((sqrt3)^2 - 1^2)` `= (3+1 - 2sqrt3)/(3-1)` `= (4-2sqrt3)/2` `2- sqrt3` answer |
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| 126. |
Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length (i) 10 cm (ii) 15 cm (iii) 21 cm |
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Answer» as we know that arc length/ radius (i) angle = `10/75 = 2/15` radian (ii) angle = `15/75 = 1/5`radian (iii)angle = `21/75 = 7/25` radian answer |
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| 127. |
Find the value of `cos^2. pi/16+cos^2. (3pi)/16+cos^2. (7pi)/16`. |
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Answer» `L.H.S.=cos^2. pi/16+cos^2.(3pi)/16+cos^2(pi/2-(3pi)/16)` `+cos^2(pi/2-pi/16)` `=cos^2.pi/16+cos^2.(3pi)/16+sin^2.(3pi)/16+sin^2.pi/16` `=(cos^2. pi/16+sin^2. pi/16)+(cos^2(3pi)/16+sin^2(3pi)/16)` `=1+1=2` |
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| 128. |
The most general value of `theta` satisfying both the equations `sin theta = 1/2: tan theta =1/sqrt3` isA. `npi+(-1)^(n)(pi)/(6), ninZ`B. `2npi+(7pi)/(6), ninZ`C. `npi-(pi)/(6), ninZ`D. `2npipm(pi)/(6), ninZ` |
| Answer» Correct Answer - B | |
| 129. |
If `tan^2theta -(1+sqrt3)tan theta + sqrt3=0`, then the general value of theta isA. `npi+(pi)/(4), npi+(pi)/(3), ninZ`B. `npi+(5pi)/(4), npi+(pi)/(3), ninZ`C. `2npi+(pi)/(4), npi+(pi)/(3), ninZ`D. `npi+(pi)/(4), 2npi+(pi)/(3), ninZ` |
| Answer» Correct Answer - A | |
| 130. |
Find the most general value of ` theta` satisfyingn the equation `tantheta=-1 and costheta=(1 )/(sqrt(2))`.A. `npi+(7pi)/(4), ninZ`B. `npi+(-1)^(n)(7pi)/(4), ninZ`C. `2npi+(7pi)/(4), ninZ`D. `2npi+(-1)^(n)(7pi)/(4), ninZ` |
| Answer» Correct Answer - C | |
| 131. |
If `tantheta+tan2theta+sqrt(3)tanthetatan2theta=sqrt(3)`, then `theta=(npi)/(3)+(pi)/(9)`.A. `(3n+1)(pi)/(9), ninZ`B. `(3n+1)(pi)/(6), ninZ`C. `(3n+1)(pi)/(3), ninZ`D. `(3n+1)(pi)/(18), ninZ` |
| Answer» Correct Answer - A | |
| 132. |
Suppose A and B are two angles such that `A , B in (0,pi)`and satisfy `sinA+sinB=1`and `cosA+cosB=0.`Then the value of `12cos2A+4cos2B`is____A. 4B. 6C. 8D. 12 |
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Answer» Correct Answer - C Since cos A + cos B = 0 `rArr A+B=pi` `therefore B = pi -A` `therefore` from sin A + sin B = 1 `sin A + sin(pi -A)=1` `rArr sin A =(1)/(2)` `therefore A=30^(@)` and `B=150^(@)` or `A=150^(@)` and `B = 30^(@)` `therefore 12 cos 60^(@)+4cos 300^(@)=8` |
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| 133. |
The expression `3[sin^4(3/2pi-alpha)+sin^4(3pi+alpha)]-2[sin^6(1/2pi+alpha)+sin^6(5pi-alpha)]` is equal to |
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Answer» Correct Answer - B `3[sin^4(3/2pi-alpha)+sin^4(3pi+alpha)]` `-2[sin^6(1/2pi-alpha)+sin^6(5pi-alpha)]` `=3(cos^4alpha+sin^4alpha)-2(cos^6alpha+sin^6alpha)` `3(1-2sin^2alphacos^2alpha)-2[(sin^2alpha+cos^2alpha)^3` `-3sin^2alphacos^2alpha(sin^2alpha+cos^2alpha)]` `=3(1-2sin^2alphacos^2alpha)-2[1-3sin^2alphacos^2alpha]` =1 |
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| 134. |
Find the values of the trigonometric function `cot``(-(15pi)/4)` |
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Answer» As `cot(-theta) = -cot theta` `:. cot(-(15pi)/4) = -cot((15pi)/4)` `=-cot(4pi-pi/4)` `=-cot(2*2pi-pi/4)` `=-cot(-pi/4)` (As `cot(2npi+theta) = cot theta`) `=cot(pi/4) = 1` `:. cot(-(15pi)/4) = 1` |
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| 135. |
The value of `cot((7pi)/16)+2cot((3pi)/8)+cot((15pi)/16)`is(a)`4`(b)` 2`(c) `-2`(d) `-4` |
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Answer» `cot((7pi)/16)+2cot((3pi)/8)+cot((15pi)/16)` `=tan(pi/2-(7pi)/16)+2tan(pi/2-(3pi)/8)+cot(pi-pi/16)` `=tan(pi/16) + 2tan(pi/8) - cot(pi/16)` `=(tan(pi/16)-cot(pi/16))+2tan(pi/8)` `=(sin(pi/16)/cos(pi/16)-cos(pi/16)/sin(pi/16))+2tan(pi/8)` `=((sin^2(pi/16)-cos^2(pi/16))/(sin(pi/16)cos(pi/16)))+2tan(pi/8)` `=(-cos(pi/8))/(1/2sin(pi/8))+2tan(pi/8)` `=2tan(pi/8)-2cot(pi/8)` `=2(sin(pi/8)/cos(pi/8) - cos(pi/8)/sin(pi/8))` `=2((sin^2(pi/8)-cos^2(pi/8))/(sin(pi/8)cos(pi/8)))` `=(2(-cos(pi/4)))/(1/2sin(pi/4))` `=-4cot(pi/4)` `=-4` So, option `d` is the correct option. |
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| 136. |
Value of expression `sin(pi/9)+sin((2pi)/9)+sin((3pi)/9)+...+sin((17pi)/9)=` |
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Answer» Correct Answer - A `sin.(pi)/(9)+sin.(2pi)/(9)+sin.(3pi)/(9)+……..+sin.(17pi)/(9)` `{:(+(sin.(2pi)/(9)+sin.(16pi)/(9))),(" ..."),(" ..."),(+(sin.(8pi)/(9)+sin.(10pi)/(9))),(" "+sin.(9pi)/(9)):}` = 0 |
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| 137. |
When `theta=(17pi)/3`, then the value of`(tan theta-cot theta)` is: (a) `2/sqrt3` (2) `-2/sqrt3` (3) `1/(2sqrt3)` (d) None |
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Answer» `theta=17/3pi=(5+2/3)pi` `=6pi-pi/3` `tantheta-cottheta=tantheta-1/tantheta` `tan(6pi-pi/3)-1/tan(6pi-pi/3)` `tan(-pi/3)-1/tan9-pi/3)` `-sqrt3-1/-sqrt3` `2/sqrt3` Option B is correct. |
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| 138. |
`(tan) (11pi)/3 (-2 sin) (4pi)/6 - 3/4 (cosec^2) pi/4 +(4cos^2) (17pi)/6 = (3-4sqrt3)/2` |
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Answer» `tan((11pi)/3) = tan((9pi+2pi)/3) = tan(3pi+(2pi)/3) = tan(pi+(2pi)/3) = tan((2pi)/3) = -sqrt3` `sin((4pi)/6) = sin((2pi)/3) = sqrt3/2` `cosec^2(pi/4) = (sqrt 2)^2 = 2` `cos^2((17pi)/6) = cos^2((18pi-pi)/6) = cos^2(3pi-pi/6) = cos^2(pi-pi/6) = (-sqrt3/2)^2 = 3/4` So, putting these values in the given expression, `-sqrt3-2*sqrt3/2-3/4*2+4*3/4 = -2sqrt3+3/2 = 1/2(3-4sqrt3) ` which is the desired value. |
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| 139. |
State if the given pairs of angles are coterminal. (a) `-185^@,535^@" (b) "1000^@,270^@" (c ) "(15pi)/4,-(17pi)/4` |
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Answer» Correct Answer - (a) coterminal (b) not coterminal (c ) coterminal (a) `alpha=-185^@,beta=535^@` `beta-alpha=535^@-(-185^@)=270^@` Hence, given angles are coterminal. (b) `alpha=1000^@,beta=270^@` `alpha-beta=1000^@-270^@=730^@` Hence, given angles are not coterminal. (c ) `alpha=(15pi)/4,beta=-(17pi)/4` Here `alpha-beta=(15pi)/4-(-(17pi)/4)=(32pi)/4=8pi=4(2pi)` Hence, given angles are coterminal. |
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| 140. |
State if the given angles are coterminal.`alpha=185^0,beta=-545^0``alpha=(17pi)/(36),beta(161pi)/(36)` |
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Answer» (i) `alpha=185^@,beta-545^@` Hence, `alpha-bet=185^@-(-545^@)=730^@` Hence, the given angles are not coterminal. `(ii)a=(17pi)/36,beta=(161pi)/36` Hence, `beta-alpha=(161pi)/36-(17pi)/36=(144pi)=36=4pi` Hence, the given angles are coterminal. |
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| 141. |
If `tanbeta=(tanalpha+tangamma)/(1+tanalphatangamma)dot`prove that `sin2beta=(sin2alpha+sin2gamma)/(1+sin2alphasin2gamma)`. |
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Answer» `tanbeta = (tanalpha+tan gamma)/(1+tanalphatangamma)` `=>tan beta = (sinalpha/cosalpha+singamma/cosgamma)/(1+sinalpha/cosalpha*singamma/cosgamma)` `=(sinalphacosgamma + singammacosalpha)/(cosalphacosgamma+sinalphasingamma)` `=>tan beta=(sin(alpha+gamma))/(cos(alpha-gamma))` Now, `sin2beta = (2tanbeta)/(1+tan^2beta)` `=(2(sin(alpha+gamma))/cos(alpha-gamma))/(1+sin^2(alpha+gamma)/cos^2(alpha-gamma) ` `=(2sin(alpha+gamma)cos(alpha-gamma))/(cos^2(alpha-gamma) + sin^2(alpha+gamma))` `=(sin(alpha+gamma+alpha-gamma) + sin(alpha+gamma-alpha+gamma))/(((1+cos2(alpha-gamma))/2)+((1-cos2(alpha+gamma))/2))` `=(sin2alpha+sin2gamma)/(1+1/2(cos(alpha-gamma) - cos(alpha+gamma)))` `=(sin2alpha+sin2gamma)/(1+1/2(2sin2alphasin2gamma)` `=(sin2alpha+sin2gamma)/(1+sin2alphasin2gamma) = R.H.S.` |
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| 142. |
If `cosalpha+cosbeta=0=sinalpha+sinbeta`, then prove that `cos2alpha +cos2beta=-2cos(alpha +beta)`. |
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Answer» Given that, `" "cosalpha+cosbeta=0=sinalpha+sinbeta` `rArr" "(cosalpha+cosbeta)^(2)-(sinalpha+sinbeta)^(2)=0` `rArr" "cos^(2)alpha-sin^(2)beta+2cosalphacosbeta-sin^(2)alpha-sin^(2)beta-2sinalphasinbeta= 0` `rArr" "cos^(2)alpha-sin^(2)alpha+cos^(2)beta-sin^(2)beta=2(sinalphasinbeta-cosalphacosbeta)` ` rArr" "cos2alpha+cos2beta=-2cos(alpha+beta)" "` Hence proved. |
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| 143. |
If `cosx-sinalphacotbetasinx=cosa ,`then the value of `tan(x/2)`is(a)`-tan(alpha/2)cot(beta/2)`(b) `tan(alpha/2)tan(beta/2)`(c)`-cot((alphabeta)/2)tan(beta/2)`(d) `cot(alpha/2)cot(beta/2)` |
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Answer» `cosx-sinalphacotbetasinx = cosalpha` `=>(1-tan^2(x/2))/(1+tan^2(x/2)) - sinalphacotbeta((2tan(x/2))/(1-tan^2(x/2))) = cosalpha` `=>1-tan^2(x/2) - sinalphacotbeta(2tan(x/2)) = cosalpha - cosalphatan^2(x/2)` `=>tan^2(x/2)(1+cosalpha) +2sinalphacotbetatan(x/2) - (1-cosalpha) = 0` `=>tan^2(x/2)(1+cosalpha) +2sinalpha(cosbeta/sinbeta)tan(x/2) - (1-cosalpha) = 0` `=>tan^2(x/2)(1+cosalpha) +2sinalpha((cos^2(beta/2)-sin^2(beta/2))/(2sin(beta/2)cos(beta/2)))tan(x/2) - (1-cosalpha) = 0` `=>tan^2(x/2)(1+cosalpha) +sinalpha(cot(beta/2)-tan(beta/2))tan(x/2) - (1-cosalpha) = 0` `=>(tan(x/2)+cot(beta/2)tan(alpha/2))(tan(x/2)-tan(beta/2)tan(alpha/2)) = 0` `=>(tan(x/2)+cot(beta/2)tan(alpha/2)) = 0 or (tan(x/2)-tan(beta/2)tan(alpha/2)) = 0` `=>tan(x/2) = - cot(beta/2)tan(alpha/2) or tan(x/2) = tan(beta/2)tan(alpha/2)` So, options `(a)` and `(b)` are the correct options. |
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| 144. |
If `"cot"(alpha+beta)=0,`then `"sin"(alpha+2beta)`can be(a)`-sinalpha`(b) `sinbeta`(c) `cosalpha`(d) `cosbeta` |
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Answer» `cot(alpha+beta) = 0` `=>alpha+beta = pi/2` Now, `sin(alpha+2beta) = sin(alpha+beta+beta)` `=sin(pi/2+beta)` `=cosbeta` `:. sin(alpha+2beta) = cos beta` So, option `(d)` is the correct option. |
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| 145. |
If `cosalpha+cosbeta=(3)/(2)and"sin"alpha+sinbeta=(1)/(2)andtheta` is the arithmetic mean of `alphaandbeta` , then `sin2theta+cos2theta` is equal toA. `(3)/(5)`B. `(7)/(5)`C. `(4)/(5)`D. `(8)/(5)` |
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Answer» Correct Answer - B `cosalpha+cosbeta=(3)/(2)andsinalpha+sinbeta=(1)/(2)` `rArr2cos((alpha+beta)/(2))cos((alpha-beta)/(2))=(3)/(2)` and `2sin((alpha+beta)/(2))cos((alpha-beta)/(2))=(1)/(2)` `rArrtan((alpha+beta)/(2))=(1)/(3)` `becausetheta=(alpha+beta)/(2)` [Given] `rArr2theta=alpha+beta` `thereforesin2theta+cos2theta=sin(alpha=beta)+cos(alpha+beta)` `=(2tan((alpha+beta)/(2)))/(1+tan^(2)((alpha+beta)/(2)))+(1-tan^(2)((alpha+beta)/(2)))/(1+tan^(2)((alpha+beta)/(2)))` `[becausesin2theta=(2tantheta)/(1-tan^(2)theta),cos2theta=(1-tan^(2)theta)/(1+tan^(2)theta)]` `=(2((1)/(3)))/(1+((1)/(3))^(2))+(1-((1)/(3))^(2))/(1+((1)/(3))^(2))=(2)/(3)xx(9)/(10)+(8)/(9)xx(9)/(10)=(6)/(10)+(8)/(10)=(7)/(5)` |
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| 146. |
If `cosalpha=1/2(x+1/x)` `cosbeta=1/2(y+1/y)` then `cos(alpha-beta)` is equal toA. `sin(alpha+beta+gamma)=singammaAAgammainR`B. `cosalphacosbeta=1AAalpha,beta inR`C. `(cosalpha+cosbeta)^2=4AAalpha,beta in R`D. `sin (alpha+beta+gamma)=sinalpha+sinbeta+singammaAAa,b,gammainR` |
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Answer» Correct Answer - A::B::C::D `cosalpha=1/2(x+1/x)andcosbeta=1/2(y+1/y)` since `xygt0`,we have `x+1/2ge2orle-2andy+1/yge2orle-2` `rArrcosalpha=1,cosbeta=1` `or cosalpha=-1,cosbeta=-1` `:. cosalphacosbeta=1` `rArralpha+beta" is an even multiple of " pi` `(cosalpha+cosbeta)^2=4` `rArr sin(alpha+beta+gamma)=sin(2npi+gamma)=singamma` Also, `sinalpha=sinbeta=0` |
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| 147. |
For `0ltphiltpi//2," if" x=sum_(n=0)^(oo) cos^(2n)phi,y=sum_(n=0)^(oo) sin^(2m)phi,z=sum_(n=0)^(oo)cos^(2n)phisin^(2n)phi`,thenA. `xyz=xz+y`B. `xyz=xy+z`C. `xyz=x+y+z`D. `xyz=yz+x` |
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Answer» Correct Answer - B::C All are infinte geometric progression with common ratio lt 1 `x=1/(1-cos^2phi)=1/sin^2phi,y=1/(1-sin^2phi)=1/cos^2phi`, `z=1/(1-cos^2phisin^2phi)` Now, `xy+z=1/(sin^2phicos^2phi)+1/(1-sin^2phicos^2phi)` `=1/(sin^2phicos^2phi(1-sin^2phicos^2phi))` `or xy+z=xyz ...(i)` Clearly, `x+y=(sin^2phi+cos^2phi)/(sin^2phicos^2phi)=xy` `:. x+y+z=xyz` [using Eq. (i)] |
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| 148. |
Four numbers `n_1,n_2,n_3andn_4` are given as `n_1=sin15^@-cos15^@,n_2=cos93^@+sin93^@,n_3=tan27^@-cot27^@,n_4=cot127^@+tan127^@`,ThenA. `n_1lt0`B. `n_2lt0`C. `n_3lt0`D. `n_4lt0` |
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Answer» Correct Answer - A::C::D `n_1=sin15^@-cos15^@lt-ve" "(cos15^@gtsin15^@)` `n_2=cos93^@+sin93^@` `=-sin3^@[email protected]" "(cos3^@gtsin3^@)` `n_3=tan27^@-cot27^@lt0" "(tan27^@[email protected])` `n_4=cot127^@+tan127^@lt0" "(tan127^@,cot127^@lt0)`. |
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| 149. |
For each natural number `nge2`, prove that `sinx_1cosx_2+sinx_2cosx_3+…+sinx_ncosx_1len//2`(where `x_1,x_2,…,x_n` are arbitrary real numbers). |
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Answer» Let the required sum be `S_(n)`. We know that `(sinx_1-cosx_2)^2=(sinx_2-cosx_3)^2+...+(sinx_(n-1)-cosx_n)^2+(sinx_n-cosx_1)^2ge0` or `(sin^2x_1-cos^2x_2)+(sin^2x_2-cos^2x_3)(sin^2x_3+cos^2x_3)+...+(sin^2x_(n)-cos^2x_n)ge2S_n` `rArr nge2S_n` or `S_nlen//2` |
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| 150. |
If `piA. `2 csc alpha`B. `-2 csc alpha`C. `csc alpha`D. `-csc alpha` |
| Answer» Correct Answer - B | |