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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.
| 251. |
If `sin(alpha + beta) = 1, sin(alpha - beta) = 1/2` then `tan(alpha + 2 beta) tan(2 alpha + beta)`= |
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Answer» `sin(alpha+beta) = 1, sin(alpha-beta) = 1/2` `alpha+beta = pi/2->(1)`, ` alpha-beta = pi/6->(2)` Adding (1) and (2), `alpha+beta+alpha-beta = pi/2+pi/6` `=>2alpha = (4pi)/6` `=>alpha = pi/3` Putting value of `alpha` in (1), `pi/3+beta = pi/2` `=>beta = pi/2-pi/3 =>beta = pi/6` `:. tan(alpha+2beta)tan(2alpha+beta) = tan(pi/3+pi/3)tan((2pi)/3+pi/6)` `=tan((2pi)/3)tan((5pi)/6)` `=(-sqrt3)(-1/sqrt3) = 1` |
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| 252. |
` 2 cos^2 B - 1 = tan^2 A`, then `cosA cosB` = |
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Answer» `2cos^2B - 1 = tan^2A` `=>2cos^2B - 1 = sec^2A-1` `=>2cos^2B = sec^2A` `=>2cos^2B = 1/cos^2A` `=>cos^2Acos^2B = 1/2` `=>cosAcosB = +-1/sqrt2` |
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| 253. |
` sin^4 theta + 2 sin^2 theta ( 1 - 1/(cosec^2 theta)) + cos^4 theta =` |
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Answer» `sin^4theta+2sin^2theta(1-1/(cosec^2theta))+cos^4theta` `=sin^4theta+2sin^2theta(1-sin^2theta)+cos^4theta` `=sin^4theta+2sin^2thetacos^2theta+cos^4theta` `=(sin^2theta+cos^2theta)^2` `=1^2 = 1` So, value of the required expression is `1`. |
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| 254. |
If ` x= sin130^@ + cos130^@` then1.) x < 0 2.) x = 0 3.) x > 0 4.) x`>= 0` |
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Answer» `x = sin 130^@ + cos 130^@` `=> x= sin(90^@+40^@)+cos(90^@+40^@)` `=>x = cos40^@-sin40^@` As we know, `sin 45^@ and cos45^@` are equal and for `theta lt 45^@`, value of `cos theta gt sin theta`, `:. cos40^@-sin40^@ gt 0` `:. x gt 0` So, option `(3)` is the correct answer. |
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| 255. |
The value of `(sin50^(@))/(sin130^(@))` is ….. . |
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Answer» Here, `" "(sin50^(@))/(sin130^(@))=(sin(180^(@)-130^(@)))/(sin130^(@))` `" "=(sin130^(@))/(sin130^(@))=1` |
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| 256. |
If for real values of x, `costheta=x+(1)/(x)`, thenA. `theta` is an acute angleB. `theta` is right angleC. `theta` is an obtuse angleD. No value of `theta` is possible |
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Answer» Correct Answer - D Here, `" "costheta=x+(1)/(x)` `rArr" "costheta=(x^(2)+1)/(x)` `" "x^(2)-xcostheta+1=0` For real value of x, `(-costheta)^(2)-4xx1xx1=0` `" "cos^(2)theta=4` `" "costheta=pm2` which is not possible. `" "[because-1lecosthetale1]` |
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| 257. |
If `tantheta=(a)/(b)`, then `bcos2theta+asin2theta` is equal toA. aB. bC. `(a)/(b)`D. None of these |
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Answer» Correct Answer - B Given that, `tantheta=(a)/(b)` `therefore" "bcos2theta+asin2theta=b((1-tan^(2)theta)/(1+tan^(2)theta))+a((2tantheta)/(1+tan^(2)theta))` `" "=b((1-(a^(2))/(b^(2)))/(1+(a^(2))/(b^(2))))+a(((2a)/(b))/(1+(a^(2))/(b^(2))))` `" "=b((b^(2)-a^(2))/(b^(2)+a^(2)))+(2a^(2)b)/(a^(2)+b^(2))` `" "=(b)/(a^(2)+b^(2))[b^(2)-a^(2)+2a^(2)]=((a^(2)+b^(2))b)/((a^(2)+b^(2)))` =b |
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| 258. |
If `tanalpha =(1)/(7) and tanbeta =(1)/(3) , then, cos2alpha` is equal toA. `sin2beta`B. `sin4beta`C. `sin2beta`D. `cos2beta` |
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Answer» Correct Answer - B Given that, `tanalpha=(1)/(7) and tanbeta=(1)/(3)` ltBrgt `" "cos2alpha=(1-(1)/(49))/(1+(1)/(49))=((48)/49)/((50)/(49))` `" "=(48)/(50)=(24)/(25)` `rArr" "cos2alpha=(24)/(25)" "...(i)` We know that, `" "sin4beta=(2tan2beta)/(1+tan^(2)2beta)" "…(ii)` and `" "tan2beta=(2tanbeta)/(1-tan^(2)beta)=(2xx(1)/(3))/(1-(1)/(9))` `" "=((2)/(3))/((8)/(9))=(2xx9)/(3xx8)=(3)/(4)` From Eq, (ii), `" "sin4beta=(2xx(3)/(4))/(1+(9)/(16))=((6)/(4))/((25)/(16))=(6xx16)/(4xx25)` `rArr" "sin4beta=(24)/(25)` `rArr" "sin4beta=cos2alpha` `therefore" "cos2alpha=sin4beta" "` [from Eq. (i)] |
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| 259. |
The value of `cos^(2)48^(@)-sin^(2)12^(@)` isA. `(sqrt(5)+1)/(8)`B. `(sqrt(5)-1)/(8)`C. `(sqrt(5)+1)/(5)`D. `(sqrt(5)+1)/(2sqrt(2))` |
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Answer» Correct Answer - A Given expression, `" "cos^(2)48^(@)-sin^(2)12^(@)` `" "=cos(48^(@)+12^(@))-cos(48^(@)-12^(@))` `" "=cos60^(@)*cos36^(@)` ltBrgt `" "=(1)/(2)*(sqrt(5)+1)/(4)` `" "=(sqrt(5)+1)/(8)` |
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| 260. |
Suppose ABCD (in order) is a quadrilateral inscribed in a circle. Whichof the following is/are always true?`secB=secD`(b) `cotA+cotC=0``cos e cA=cos e cC`(d) `tanB+tanD=0` |
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Answer» `A+C=180` `B+D=180` `B=180-D` `secB=sec(180-D)` `secB=-secD` `A=180-C` `cotA=cot(180-C)=-cotC` `cotA+cotC=0` `A=180-C` `cosecA=cosec(180-C)=cosecC` `B=180-D` `tanB=tan(180-D)=-tanD` `tanB+tanD=0`. |
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| 261. |
Which of the following is not possible?A. `sintheta=5/3`B. `tantheta=1002`C. `costheta=(1+p^2)/(1-p^2),(pne0,pm1)`D. `sectheta=1/2` |
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Answer» (b) `sintheta=5/3" is not possible as "-1lesinthetale1`. `costheta=(1+p^2)/(1-p^2)" is not possible, as in "(1+p^2)/(1-p^2)" the numerator is always greater than the denominator for any value of p other than p = 0. Hence, " (1+p^2)/(1-p^2) " does not lie in "[-1, 1]`. `sectheta=1/2` is not possible as `sectheta in(-oo,-1]uu[1,oo).tantheta=1002` is possible as `tantheta`can take any real value. |
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| 262. |
Let (-3, -4) be a point on the terminal side of `theta`. Find the sine, cosine and tangent of `theta`. |
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Answer» Correct Answer - `sintheta=-4/5,costheta=-3/5,tantheta=4/3` Here `x=-3,y=-4` Terminal angle `theta`lies in the third quadrant. Now, `r=sqrt(x^2+y^2)=sqrt((-3)^2+(-4)^2)=sqrt25=5` `:. sintheta =y/r=-4/5` `costheta =x/r=-3/5` `tantheta =y/x=4/3` |
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| 263. |
If `angleA=90^(@)` in the `DeltaABC` , then `tan^(-1)((c)/(a+b))+tan^(-1)((b)/(a+c))` is equal to |
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Answer» Correct Answer - C `becauseDeltaABC` right angled at A. `thereforea^(2)=b^(2)+c^(2)` Now , `tan^(-1)((c)/(a+b))+tan^(-1)((b)/(a+c))` `=tan^(-1)[((c)/(a+b)+(b)/(a+c))/(1-((c)/(a+b))((b)/(a+c)))]` `=tan^(-1)[(ac+c^(2)+ab+b^(2))/(a^(2)+ac+ab+bc-bc)]` `=tan^(-1)[(a^(2)+ac+ab)/(a^(2)+ac+ab)]` [using Eq . (i)] `=tan^(-1)(1)=(pi)/(4)` |
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| 264. |
In `DeltaABC,(a-b)^(2)"cos"^(2)(C)/(2)+(a+b)^(2)"sin"^(2)(C)/(2)` is equal toA. `b^(2)`B. `c^(2)`C. `a^(2)`D. `a^(2)+b^(2)+c^(2)` |
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Answer» Correct Answer - B `(a-b)^(2)"cos"^(2)(C)/(2)+(a+b)^(2)"sin"^(2)(C)/(2)` `=(a^(2)+b^(2)-2ab)"cos"^(2)(C)/(2)+a^(2)+b^(2)+2ab"sin"^(2)(C)/(2)` `=a^(2)("cos"^(2)(C)/(2)+"sin"^(2)(C)/(2))+b^(2)("cos"^(2)(C)/(2)+"sin"^(2)(C)/(2))-2ab("cos"^(2)(C)/(2)-"sin"^(2)(C)/(2))` `=a^(2)xx1+b^(2)-2ab((s(s-c))/(ab)-((s-a)(s-b))/(ab))` `=a^(2)+b^(2)-(2ab)/(ab)[((a+b+c)/(2))((a+b+c)/(2)-c)` `-((a+b+c)/(2)-a)xx((a+b+c)/(2)-b)` `=a^(2)+b^(2)-(2)/(4)[(a+b+c)(a+b-c)-(b+c-a)(a+c-b)]` `=a^(2)+b^(2)-(1)/(2)[(a+b^(2))-c^(2)-(ab+bc-b^(2)+ac+c^(2)-bc-a^(2)-ac+ab)]` `=a^(2)+b^(2)-(1)/(2)[a^(2)+b^(2)+2ab-c^(2)+a^(2)+b^(2)-c^(2)-2ab]` `=a^(2)+b^(2)-(1)/(2)[2a^(2)+2ab^(2)-2c^(2)]` `=a^(2)+b^(2)-[a^(2)+b^(2)-c^(2)]=c^(2)` |
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| 265. |
The value of `tan^-1(1/3)+tan^-1(2/9)+tan^-1(4/33)+tan^-1(8/129)+....n` terms is: |
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Answer» `T_r=tan^(-1)((2^(r-1))/(2^(2r-1)+1))` `r_1=1/3` `r_2=2/9` `r_3=4/33` `2^(r-1)=2*2^(r-1)-2^(r-1)=2^r-2^(r-1)` `T_r=tan^(-1)((2^r*2^(r-1))/(1+2^r*2^(r-1)))` `T_1=tan^(-1)(2)-tan(1)` `T_2=tan^(-1)(4)-tan^(-1)(2)` `T_3=tan^(-1)(8)-tan^(-1)(4)` . . . `T_n=tan^(-1)(2^n)-tan^(-1)(2^(n-1))` `T_1+T_2+t_3+...+T_n=tan^(-1)2^n-pi/4` option 1 is correct. |
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| 266. |
Prove that `tan(pi/12) . tan(5 pi/12) . tan(7pi/12) . tan(11pi/12) =1` |
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Answer» `tan5/12pi=cot(pi/2-5/12pi)=cotpi/12` `tan7/12pi=tan(pi-7/12pi)=-tan5/12pi=-cotpi/12` `tan7/12pi=-tanpi/12` `tanpi/12*cotpi/12-cotpi/12*-tanpi/12` 1. |
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| 267. |
If : `a * cos A-b * sin A=c, "then" : a * sin A +b* cos A=`A)`sqrt(a^(2)+b^(2)-c^(2))` B)`sqrt(a^(2)-b^(2)+c^(2))` C)`sqrt(b^(2)+c^(2)-a^(2))` D)`sqrt(b^(2)+c^(2)+a^(2))`A. `sqrt(a^(2)+b^(2)-c^(2))`B. `sqrt(a^(2)-b^(2)+c^(2))`C. `sqrt(b^(2)+c^(2)-a^(2))`D. `sqrt(b^(2)+c^(2)+a^(2))` |
| Answer» Correct Answer - A | |
| 268. |
If `sec theta-tan theta=x` and `sectheta+tan theta=1/x` then sinf `sin theta` and `cos theta`A. `(1+x^(2))/(1-x^(2))`B. `(1-x^(2))/(1+x^(2))`C. `x^(2) + (1)/(x^(2))`D. `(1-x^(2))/(1+x^(2))` |
| Answer» Correct Answer - D | |
| 269. |
If : `asec^(2) theta - b * tan^(2) theta = c,"then" = sin theta =`A)`sqrt((a +c)/(b + c)` B)`sqrt((a -c)/(b - c)` C)`sqrt((a -c)/(b + c)` D)`sqrt((a + c)/(b - c)`A. `sqrt((a +c)/(b + c)`B. `sqrt((a -c)/(b - c)`C. `sqrt((a -c)/(b + c)`D. `sqrt((a + c)/(b - c)` |
| Answer» Correct Answer - B | |
| 270. |
Let `x=sin1^0,`then the value of the expression.`1/(cos0^0dotcos1^0)+1/(cos1^0dotcos2^0)+1/(cos2^0dotcos3^0)+1/(cos44^0dotcos45^0)`is equal to(a) `x`(b) `1/x`(c) `(sqrt(2))/x`(d) `x/(sqrt(2))` |
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Answer» `1/sin1[(sin(1-0))/(cos0cos1)+(sin(2-1))/(cos2cos2)+....+(sin(45-44))/(cos44cos45)]` `1/sin1[tan1-tan0+tan2-tan2+tan3-tan2+...+tan45-tan44]` `1/sin1*[1-0]` `1/sin1` `1/x`. |
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| 271. |
`lnDeltaA B C , b^2cos2A-a^2cos2B=`A. `0`B. `a^(2)+b^(2)`C. `a^(2)-b^(2)`D. `b^(2)-a^(2)` |
| Answer» Correct Answer - D | |
| 272. |
In `triangleABC, (a-b)^(2)cos^(2)((C)/(2))+(a+b)^(2)sin^(2)((C)/(2))`A. `b^(2)`B. `c^(2)`C. `a^(2)`D. `a^(2)+b^(2)+c^(2)` |
| Answer» Correct Answer - B | |
| 273. |
In `triangleABC, (b-c)^(2)cos^(2)((A)/(2))+(b+c)^(2)sin^(2)((A)/(2))`A. `b^(2)`B. `c^(2)`C. `a^(2)`D. `a^(2)+b^(2)+c^(2)` |
| Answer» Correct Answer - A | |
| 274. |
In `triangleABC, (asin(B-C))/(b^(2)-c^(2))=`A. `sinA`B. `-sinA`C. `asinA`D. constant |
| Answer» Correct Answer - D | |
| 275. |
In `triangleABC, (sinB)/(sin(A+B))` =A. `(b)/(a+b)`B. `(b)/(c)`C. `(-b)/(c)`D. `(c)/(b)` |
| Answer» Correct Answer - B | |
| 276. |
In ` A B C ,(cot A/2+cotB/2)(asin^2B/2+bsin^2A/2)=``cotC`(b) `ccotC`(c) `cotC/2`(d) `ccotC/2`A. `cot((C)/(2))`B. `ccot((C)/(2))`C. `cotC`D. `ccotC` |
| Answer» Correct Answer - B | |
| 277. |
In any `DeltaABC`, prove that `"a sin (B - C) + bsin(C - A) + csin(A - B)=0`.A. `0`B. `a+b+c`C. `a^(2)+b^(2)+c^(2)`D. `2(a^(2)+b^(2)+c^(2))` |
| Answer» Correct Answer - A | |
| 278. |
In `triangleABC, (bsin(C-A))/(c^(2)-a^(2))=`A. `sinB`B. `-sinB`C. `bsinB`D. constant |
| Answer» Correct Answer - D | |
| 279. |
In `triangleABC, (csin(A-B))/(a^(2)-b^(2))=`A. `sinC`B. `-sinC`C. `csinC`D. constant |
| Answer» Correct Answer - D | |
| 280. |
In `DeltaABC` if the angle A, B, C are in A.P. then `(a+c)/sqrt(a^2-ac+c^2)=`A. `2cos((A-C)/(2))`B. `sin((A+C)/(2)`C. `sin(A)/(2)`D. `sin((C)/(2))` |
| Answer» Correct Answer - A | |
| 281. |
In `DeltaABC if 2s =a +b+c and (s-b)(s-c)=x" sin"^(2)(A)/(2)` ,then the value of x isA. `ab`B. `bc`C. `ac`D. `abc` |
| Answer» Correct Answer - B | |
| 282. |
In the following match each item given under the Column I to its correct answer given under the Column II. `{:(,"Column I",,"Column II"),((i),sin(x+y)sin(x-y),(a),cos^(2)x-sin^(2)y),((ii),cos(x+y)cos(x-y),(b),1-tantheta//1+tantheta),((iii),cot((pi)/(4)+theta),(c),1+tantheta//1-tantheta),((iv),tan((pi)/(4)+theta),(d),sin^(2)x-sin^(2)y):}` |
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Answer» (i) `sin(x+y)sin(x-y)=sin^(2)x-sin^(2)y` (ii) `cos(x+y)cos(x-y)=cos^(2)x-sin^(2)y` (iii) `cot((pi)/(4)+theta)=(cot""(pi)/(4)cottheta-1)/(cot""(pi)/(4)+cottheta)` `" "=(-1+cottheta)/(1+cottheta)=(1-tantheta)/(1+tantheta)` (iv) `tan((pi)/(4)+theta)=(tan""(pi)/(4)+tantheta)/(1-tan""(pi)/(4)tantheta)=(1+tantheta)/(1-tantheta)` Hence, the correct mathes are `(i)to(d), (ii)to (a), (iii)to(b), (iv)to(c)`. |
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| 283. |
`2sin^2(pi/6 )+c o s e c^2((7pi)/6)cos^2(pi/3)=3/2` |
| Answer» `L.H.S.=2sin^2(pi/6)+cosec^2((7pi)/6)cos^2(pi/3)``=2sin^2(pi/6)+cosec^2(pi+pi/6)cos^2(pi/3)``=2sin^2(pi/6)+cosec^2(-pi/6)cos^2(pi/3)``=2(1/2)^2 +(-2) ^2**1/4 = 1/2+1 = 3/2=R.H.S.` | |
| 284. |
Find the value of other five trigonometric function `cosx=-1/2,` x lies in third quadrant. |
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Answer» In the third quadrant, only `tanx` and `cotx` will be positive. Rest will be negative. Here,`cosx = -1/2` `sinx = sqrt(1-cos^2x) = sqrt(1-1/4) = sqrt(3/4) = -sqrt3/2` `tanx = sinx/cosx = sqrt3` `cotx = 1/tanx = 1/sqrt3` `secx=1/cosx = -2` `cosecx = 1/sinx = -2/sqrt3` |
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| 285. |
If `cotx=-5/(12)`, lies in second quadrant, find the values of other five trigonometric functions. |
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Answer» As `x` lies in second quadrant, only `sinx` and `cosec x` will be positive. Rest will be negative. We are given,`cotx = -5/12` So, `tanx = 1/cotx = -12/5` `secx = sqrt(1+tan^2x) = sqrt(1+144/25) = -13/5` `cosx=1/secx = -5/13` `sinx = tanxcosx = -12/5**-5/13 = 12/13` `cosec x = 1/sinx = 13/12` |
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| 286. |
`cot^2``pi/6``+cose c``(5pi)/6``+3tan^2``pi/6=6` |
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Answer» `L.H.S. = cot^2pi/6+cosec((5pi)/6)+3tan^2(pi/6)` `=cot^2pi/6+cosec(pi-pi/6)+3tan^2(pi/6)``=cot^2pi/6+cosec(pi/6)+3tan^2(pi/6)``=sqrt3^2+2+3(1/sqrt3)^2=3+2+1 = 6=R.H.S.` |
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| 287. |
If `cosx=-3/5`, x lies m the third quadrant, find the values of other five trigonometric functions. |
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Answer» As `x` lies in third quadrant, only `tanx` and `cotx` will be positive. Rest will be negative. Now, we are given `cosx=-3/5` `secx = 1/cosx = -5/3` `tanx = sqrt(sec^2x-1) = sqrt(25/9-1) = 4/3` `cotx = 1/tanx = 3/4` `sinx = tanxcosx = 4/3(-3/5) =-4/5` `cosecx =1/sinx = -5/4` |
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| 288. |
Differentiate each of the following functions with respect to `x :``sin^(-1)(2xsqrt(1-x^2)),-1/(sqrt(2))A. `2cos^(-1)x`B. `cos^(-1)x`C. `2sin^(-1)x`D. `sin^(-1)x` |
| Answer» Correct Answer - C | |
| 289. |
If `tan^(-1)x-tan^(-1)y=tan^(-1)A,` then A=A. `(x-y)/(1+xy)`B. `(x+y)/(1-xy)`C. `x-y`D. `x+y` |
| Answer» Correct Answer - A | |
| 290. |
The equation `2cos^2x/2sin^2x=x^2+x^(-2);0` |
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Answer» `(2cos^2x/2)sin^2x=x^2+1/x^2` LHS `0ltxltpi/2` `cos0>cosx>=cospi/2` `1>cosx>=0` `2>1+cosx>=1` `1<1+cosx<2` `-1<=sinx<=1` `0<=(1+cosx)(sin^2x)<2` `0<=LHS<=2` RHS `x^2+1/x^2=(x-1/x)^2+2` `0<=(x-1/x)^2 |
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| 291. |
`cos(alpha-beta)=1a n dcos(alpha+beta)=l/e ,`where `alpha,betamu in [-pi,pi]`. Number of pairs of `alpha,beta`which satisfy both the equations is0 (b)1 (c) 2(d) 4 |
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Answer» `cos(alpha-beta)=1` `cos(alpha-beta)=cos0` `alpha-beta=0` `alpha=beta` `cos(alpha+beta)=1/e` `cos2alpha=1/e` `alpha=beta=4` Option D is correct. |
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| 292. |
The principal solution `sqrt(3)cosecx-2=0` areA. `(pi)/(3), (3pi)/(3)`B. `(4pi)/(3), (5pi)/(3)`C. `(-pi)/(3), (-2pi)/(3)`D. `(pi)/(3), (4pi)/(3)` |
| Answer» Correct Answer - A | |
| 293. |
The principal solution `cosecx=-2` areA. `(-7pi)/(6), (-11pi)/(6)`B. `(pi)/(6), (7pi)/(6)`C. `(pi)/(6), (5pi)/(6)`D. `(7pi)/(6), (11pi)/(6)` |
| Answer» Correct Answer - D | |
| 294. |
The principal solution `cosecx=2` areA. `(pi)/(3), (pi)/(6)`B. `(pi)/(6), (5pi)/(6)`C. `(7pi)/(6), (11pi)/(6)`D. `(-pi)/(6), (-5pi)/(6)` |
| Answer» Correct Answer - B | |
| 295. |
The principal solution `cosecx=sqrt(2)` areA. `(pi)/(2), (2pi)/(3)`B. `(-pi)/(4), (-3pi)/(4)`C. `(5pi)/(4), (7pi)/(4)`D. `(pi)/(4), (3pi)/(4)` |
| Answer» Correct Answer - D | |
| 296. |
The principal solution `tanx=-sqrt(3)` isA. `(pi)/(3)`B. `(5pi)/(3)`C. `(4pi)/(3)`D. `-(2pi)/(3)` |
| Answer» Correct Answer - B | |
| 297. |
The principal solution `tanx=-sqrt(3)` isA. `(2pi)/(3)`B. `(pi)/(3)`C. `(4pi)/(3)`D. `(-2pi)/(3)` |
| Answer» Correct Answer - A | |
| 298. |
The principal solution `tanx=(1)/(sqrt(3))` isA. `(11pi)/(6)`B. `(7pi)/(6)`C. `(pi)/(6)`D. `(-5pi)/(6)` |
| Answer» Correct Answer - A | |
| 299. |
The principal solution `tanx=(1)/(sqrt(3))` isA. `(7pi)/(6)`B. `(pi)/(6)`C. `(5pi)/(6)`D. `(-5pi)/(6)` |
| Answer» Correct Answer - C | |
| 300. |
The principal solution `tanx=sqrt(3)` isA. `(2pi)/(3)`B. `(4pi)/(3)`C. `(-pi)/(3)`D. `(-4pi)/(3)` |
| Answer» Correct Answer - B | |