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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. |
If A, B and C are interior angles of a triangle ABC, then show that `sin((B+C)/2)=cos(A/2)`. |
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Answer» `R.H.S. = cos(A/2)` As we know, `A+B+C = 180^@` `A = 180-(B+C)` `:.cosA/2 = cos((180-(B+C))/2)=cos(90-(B+C)/2)`As we know, `cos(90-x) = sinx` `:.cosA/2 = sin((B+C)/2)=L.H.S.` |
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| 2. |
If A, B and C are interior angles of a triangle ABC, then show that `sin((B+C)/2)=cosA/2`. |
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Answer» In `deltaABC` `A+B+C=180^(@)` `rArrB+C=180^(@)-A` `rArr(B+C)/(2)=90^(@)-(A)/(2)` `rArr"sin" (B+C)/(2)=sin(90^(@)-(A)/(2))` `rArr"sin"(B+C)/(2)="cos"(A)/(2)` |
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| 3. |
Prove that : `(sinalpha+cosalpha)(tanalpha+cotalpha)=secalpha+"cosecα"`. |
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Answer» L.H.S. `=(sinalpha+cosalpha)(tanalpha+cotalpha)` `=(sinalpha+cosalpha)((sinalpha)/(cosalpha)+(cosalpha)/(sinalpha))=(sinalpha+cosalpha)((sin^(2)alpha+cos^(2)alpha)/(sinalphacosalpha))` `=(sinalpha+cosalpha)((1)/(sinalphacosalpha))=(sinalpha)/(sinalphacosalpha)+(cosalpha)/(sinalphacosalpha)` `=(1)/(cosalpha)+(1)/(sinalpha)=secalpha+"cosec"alpha=R.H.S.` |
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| 4. |
Given `sec theta=(13)/(12)`, calculate all other trigonometric ratios. |
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Answer» `sectheta=13/12` so, H=13,B=12,P=5 `costheta=B/H=12/13` `sintheta=P/H=5/13` `tantheta=P/B=5/12` `cosectheta=H/P=13/5` `cottheta=B/P=12/5` |
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| 5. |
Evaluate : `cos60^(@)cos30^(@)+ sin 60^(@)sin30^(@)` |
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Answer» We know that , `cos60^(@)=(1)/(2),cos30^(@)=(sqrt(3))/(2),sin 60^(@) = (sqrt(3))/(2),sin=(1)/(2)` `:. cos60^(@) cos 30^(@) +sin 60^(@) sin 30^(@)` `=(1)/(2)xx(sqrt(3))/(2)+(sqrt(3))/(2)xx(1)/(2)=(sqrt(3)+sqrt(3))/(4)=(2sqrt(3))/(4)=(sqrt(3))/(2)` |
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| 6. |
Show that : sin `30^(@)=sqrt((1-cos60^(@))/(2))` |
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Answer» `sin x=1` `rArrsin x=sin 90^(@) " " rArr " " x=90^(@)` `rArr(x)/(3)=(90^(@))/(3)=30^(@) " " :." " tan""(x)/(3)=tan30^(@)=(1)/(sqrt(3))` |
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| 7. |
given that `tan(theta_1 + theta_2) = (tantheta_1 + tantheta_2)/(1-tantheta_1*tantheta_2)` find `(theta_1+theta_2)` when `tantheta_1 =1/2 , tantheta_2=1/3` |
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Answer» `tan(theta_(1)+theta_(2))=(tantheta_(1)+tantheta_(2))/(1-tantheta_(1)*tantheta_(2))= (1/(2)+(1)/(3))/(1-(1)/(2)xx(1)/(3))=((3+2)/6)/(1-(1)/(6))=((5)/6)/(5/(6))=1=tan45^(@)` `rArr theta_(1)+theta_(2)=45^(@)` |
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| 8. |
If `asintheta+bcostheta=c` then prove that `acostheta-bsintheta=sqrt(a^2+b^2-c^2)` |
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Answer» `asintheta+bcostheta=c` `rArr(asintheta+bcostheta)^(2)=c^(2)` `rArra^(2)sin^(2)theta+b^(2)cos^(2)theta+2ab "sin"thetacostheta=c^(2)` `rArra^(2)+b^(2)-c^(2)=a^(2)cos^(2)theta+b^(2)sin^(2)theta-2ab"sin thetacostheta` `rArra^(2)+b^(2)-c^(2)=(acostheta-bsintheta)^(2)` `rArracostheta-bsintheta=sqrt(a^(2)+b^(2)-c^(2))` |
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| 9. |
Shown that : `tan48^(@)tan23^(@)tan42^(@)tan67^(@)=1` |
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Answer» L.H.S. `=tan48^(@)tan23^(@)tan42^(@)tan67^(@)` `=tan48^(@)tan23^(@)tan(90^(@)-48^(@))tan(90^(@)-23^(@))` `=tan48^(@)tan23^(@)cot48^(@)cot23^(@)` `=tan48^(@)tan23^(@)*(1)/(tan48^(@))*(1)/(tan23^(@))=1=R.H.S`. |
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| 10. |
Express `cot 85^o +cos75^o`in terms of trigonometric ratios of angles between `0^o`and `45^o` |
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Answer» Here, we will use `cotx = tan(90-x)` `cosx = sin(90-x)` Now, our expression is, `cot85^@+cos75^@ = tan(90^@-85^@)+sin(90^@-75^@)` `tan5^@+sin15^@` |
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| 11. |
Evaluate : (i)`tan42^(@)-cot48^(@)` (ii)`sec36^(@)-"cosec"54^(@)` |
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Answer» (i) `tan42^(@)-cot48^(@)=tan42^(@)-cot(90^(@)-42^(@))` (ii) `sec36^(@)-"cosec " 54^(@)=sec36^(@)-"cosec"(90^(@)=36^(@))` `=sec36^(@)-sec36^(@)=0` |
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| 12. |
If `0^(@)anglethetaangle90^(@)`, then find the value of `theta` from the equation `(cos^(2)theta)/(cot^(2)theta-cos^(2)theta)=3`. |
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Answer» `(cos^(2)theta)/(cot^(2)theta-cos^(2)theta)=3` `rArr cos^(2)theta/((cos^(2)theta)/(sin^(2)theta)-cos^(2)theta)=3rArr(cos^(2)theta)/(cos^(2)theta((1)/(sin^(2)theta)-))=3` ` rArr(1)/(1/(sin^(2)theta)-1)=3rArr(1/(1-sin^(2)theta)=3)/(sin^(2)theta)` `rArr(sin^(2)theta)/(1-sin^(2)theta)=3rArr(sin^(2)theta)/(cos^(2)theta)=3` `rArrtan^(2)theta=(sqrt(3))^(2)` `rArrtantheta=sqrt(3)` (taking positive sign only) `rArrtantheta=tan60^(@)` `rArrtheta=60^(@)` |
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| 13. |
Evaluate the following : (i)`(sin58^(@))/(cos32^(@))` (ii) `(sec42^(@))/("cosec"48^(@))` (iii)`(tan37^(@))/(cot53^(@))` |
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Answer» (i) `(sin58^(@))/(cos32^(@)) =(sin(90^(@)=32))/(cos32^(@))=(cos32^(@))/(cos32^(@))=1` (ii) `(sec42^(@))/("cosec"48^(@))=(sec(90^(@)-48))/("cosec"48^(@))=("cosec"48^(@))/("cosec"48^(@))=1` (iii) `(tan37^(@))/(cot53^(@))=(tan(90^(@)-53^(@)))/(cot53^(@))=(cot53^(@))/(cot53^(@))=1` |
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| 14. |
Express each of the following in terms of trigonometric ratios of angles between `0^(@)and 45^(@)` (i) `sin70^(@)+sec70^(@)` " " (ii) `tan65^(@)+"cosec "65^(@)` " " (iii) ``cos81^(@)+cot80^(@)` |
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Answer» (i) `sin70^(@)+sec70^(@)=sin(90^(@)-20^(@))+sec(90^(@)-20^(@))=cos20^(@)+"cosec " 20^(@)` (ii) `tan65^(@)+"cosec " 65^(@)=tan(90^(@)-25^(@))+"cosec"(90^(@)-25^(@))=cot25^(@)+sec25^(@)` (iii) `cos81^(@)+cot80^(@)=cos(90^(@)=9^(@))+cot(90^(@)-10^(@))=sin9^(@)+tan10^(@)` |
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| 15. |
If `2sin^2theta-cos^2theta=2,` find the value of `theta.` |
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Answer» Given, `2sin^(2)theta-cos^(2)theta=2` `rArr 2sin^(2)theta-(1-sin^(2)theta)=2` `[therefore sin^(2)theta+ cos^(2)theta=1]` `rArr 2sin^(2)theta + sin^(2)theta-1=2` `rArr 3sin^(2)theta=3` `rArr sin^(2)theta=` `rArr sintheta=1=sin90^(@)` `[therefore sin 90^(@)=1]` `therefore theta=90^(@)` |
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| 16. |
Without using trigonometric tables , evaluate : `((tan20^(@))/(cosec70^(@)))^(2)+((cot20^(@))/(sec70^(@)))^2+2tan 15^(@)tan37^(@)tan53^(@)tan60^(@)tan75^(@)` |
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Answer» `((tan20^(@))/("cosec"70^(@)))^(2)+((cot20^(@))/(sec70^(@)))=+2tan 15^(@)tan37^(@)tan53^(@)60^(@)tan75^(@)` `{(tan20^(@))/("cosec"(90^(@)-20^(@)))}^(2)+{(cot20^(@))/(sec(90^(@)-20^(@)))}` `+2tan15^(@)tan37^(@)tan(90^(@)-37^(@))*(sqrt(3))tan(90^(@)-15^(@))` `((tan20^(@))/(sec20^(@)))^(2)+((cot20^(@))/("cosec "20^(@)))^(2)+2tan15^(@)tan37^(@)cot37^(@)*(sqrt(3))cot15^(@)` `=((sin20^(@)//cos20^(@))/(1//cos20^(@)))^(2)+((cos20^(@)//sin20^(@))/(1//sin20^(@)))^(2)+2sqrt(3)tan15^(@)tan37^(@)*(1)/(tan37^(@))*(1)/(tan15^(@))` `sin^(2)20^(@)+cos^(2)20^(@)+2sqrt(3)=1+2sqrt(3)` |
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| 17. |
Prove the following identity: `sec^4theta-sec^2theta=tan^4theta+tan^2theta` |
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Answer» L.H.S. `=tan^(4)theta+tan^(2)theta=tan^(2)theta(tan^(2)theta+1)` `=(sec^(2)theta-1)sec^(2)theta=sec^(4)theta-sec^(2)theta=R.H.S.` |
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| 18. |
Without using trigonometric tables , evaluate the following : `(cot(90^(@)-theta)*sin(90^(@)-theta))/(sintheta)+(cot40^(@))/(tan50^(@))-(cos^(2)20+cos^(2)70^(@))` |
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Answer» `(cot(90^(@)-theta)*sin(90^(@)-theta))/(sintheta)+(cot40^(@))/(tan50^(@))-(cos^(2)20^(@)+cos^(2)70^(@))` `=(tan*costheta)/(sintheta)+(cot(90^(@)-50^(@)))/(tan50^(@))-{cos^(2)(90^(@)-70^(@))+cos^(2)70^(@)}` `=(sintheta*costheta)/(costheta*sintheta)+(tan50^(@))/(tan50^(@))-(sin^(2)70^(@)+cos^(2)70^(@))` `=1+1-1=1` |
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| 19. |
If `cos9alpha=sinalpha` and `9alpha lt 90^(@)`, then the value of `tan5alpha` isA. `1/sqrt(3)`B. `sqrt(3)`C. 1D. 0 |
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Answer» Correct Answer - C Given, cos`9alpha=sinalpha` and `9alpha lt90^(@)` i.e., acute angle. `sin(90^(@)-9alpha)=sinalpha` `[therefore cosA= sin(90^(@)-A)]` `rArr 90^(@)-9alpha=alpha` `rArr alpha=9^(@)` `therefore tan5alpha=tan(5 xx 9^(@))= tan45^(@)=1` `[therefore tan45^(@)=1]` |
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| 20. |
If `sinalpha=1/2` and `cosbeta=1/2`, then the value of `(alpha+beta)` isA. `0^(@)`B. `30^(@)`C. `60^(@)`D. `90^(@)` |
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Answer» Correct Answer - D Given, `sinalpha=1/2=sin30^(@)` `[therefore sin 30^(@)=1/2]` `rArr alpha=30^(@)` and `cosbeta=1/2=cos60^(2)` `[therefore cos60^(@)=1/2]` `rArr beta=60^(@)` `therefore alpha+beta=30^(@)+60^(@)=90^(@)` |
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| 21. |
If cos `(alpha+beta)=0`, then `sin(alpha-beta)`canbe reduced to : (a) `cosbeta` (b)`cos2beta` ( c) `sinalpha` (d) `sin2alpha` |
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Answer» Given that , `cos(alpha+beta)=0=cos90^(@)` `:. alpha+beta=90^(@)` `:.alpha=90^(@)-beta` `:.sin (alpha-beta)=sin[(90^(@)-beta)]=sin(90^(@)-2beta)=cos2beta` |
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| 22. |
Show that `1+cot^2alpha/(1+cosecalpha)=cosec alpha` |
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Answer» LHS = `1+ (cot^(2)alpha)/(1+"cosec"alpha)=1+(cos^(2)alpha//sin^(2)alpha)/(1+1//sinalpha)` `[therefore cottheta=(costheta)/(sintheta)` and `"cosec"theta=1/(sintheta)]` `=1+ (cos^(2)alpha)/(sinalpha(1+sin alpha))` `=(sinalpha + 1)/(sinalpha(sinalpha+1))=1/sinalpha` `[therefore "cosec"theta=1/sintheta]` `="cosec"alpha=RHS` |
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| 23. |
If `sintheta + 2 costheta=1`,then prove that `2sintheta-costheta=2`. |
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Answer» Given, `sintheta+2costheta=1` On squaring both sides, we get `(sintheta+2costheta)^(2)=1` `sin^(2)theta+4cos^(2)theta+4sintheta.costheta=1` `rArr (1-cos^(2)theta) + 4(1-sin^(2)theta)+4sintheta.costheta=1` `[therefore sin^(2)theta+ cos^(2)theta=1]` `rArr 4sin^(2)theta + cos^(2)theta-4sintheta. costheta=4` `rArr (2sintheta-costheta)^(2)=4` `[therefore a^(2)+b^(2)-2ab=(a-b)^(2)]` `rArr 2sintheta-costheta=2` Hence proved. |
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| 24. |
If `cos(alpha + beta)=0`, then sin`(alpha-beta)` can be reduced toA. `cosbeta`B. `cos2beta`C. `sinalpha`D. `sin2alpha` |
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Answer» Correct Answer - B Given, `cos(alpha+beta)=0=cos 90^(@)` `[therefore cos90^(@)=0]` `rArr alpha+beta=90^(@)` `rArr alpha=90^(@)- beta`……………..(i) Now, `sin(alpha-beta) = sin(90^(@)-beta-beta)` [put the value from Eq.(i)] `=sin(90^(@)-2beta)` `=cos2beta` `[therefore sin(90^(@)-theta)=costheta]` Hence, `sin(alpha-beta)` can be reduced to `cos2beta`. |
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| 25. |
If `sintheta=a/b`, then `costheta` is equal toA. `b/(sqrt(b^(2)-a^(a))`B. `b/a`C. `(sqrtb^(2)-a^(2))/b`D. `a/(sqrt(b^(2)-a^(2))` |
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Answer» Correct Answer - C Given, `sintheta=a/b` `[therefore sin^(2)theta+ cos^(2)theta=1 rArr costheta=sqrt(1-sin^(2)theta)]` `therefore costheta=sqrt(1-sin^(2)theta)` `=sqrt(1-(a/b)^(2)) = sqrt(1-a^(2)/b^(2))= (sqrtb^(2)-a^(2))/b` |
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| 26. |
Prove that : `(secA+1)/(tanA)=(tanA)/(secA-1)` |
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Answer» L.H.S. `=(secA+1)/(tanA)=(tanA+1)/(secA)xx(secA-1)/(secA-1)` [divide numberator and denominator by (secA-1)] `=(sec^(2)A-1)/(tanA(secA-1))=(tan^(2)A)/(tanA (secA-1))=(tanA)/(secA-1)=R.H.S` |
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| 27. |
If `4 tantheta=3,` then `((4sintheta-costheta)/(4 sintheta+costheta))` is equal toA. `2/3`B. `1/3`C. `1/2`D. `3/4` |
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Answer» Correct Answer - C Given, 4tan`theta=3` `rArr tantheta=3/4` `therefore (4sintheta-costheta)/(4sintheta-costheta) = (4(sintheta)/(costheta)-1)/(4(sintheta)/(costheta)+1)` [divide by cos`theta` in both numerator and denominator] `(4tantheta-1)/(4tantheta+1)` `[therefore tantheta=(sintheta)/(costheta)]` `=(4(3//4)-1)/(4(3//4)+1)=(3-1)/(3+1)=2/4=1/2` [put the value from Eq. 9i)]` |
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| 28. |
if `tantheta+sectheta=l ` then prove that `tantheta=(l^2+1)/(2l)` |
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Answer» Given, `tantheta+sectheta=l` [multiply by `(sectheta-tantheta)` on numerator and denominator LHS] `((tantheta+sectheta)(sectheta-tantheta))/((sectheta-tantheta))=l rArr (sec^(2)theta-tan^(2)theta)/(sectheta-tantheta)=l` `1/(sectheta-tantheta)=l` `[therefore sec^(2)theta - tan^(2)theta=1]` `rArr sectheta-tantheta=1/l` ...............(ii) On adding Eqs. (i) and (ii), we get `2sectheta= l + 1/l` `sectheta=(l^(2)+1)/(2l)` Hence Proved. |
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| 29. |
Prove that `(1+sectheta-tantheta)/(1+sectheta+tantheta) = (1-sintheta)/(costheta)` |
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Answer» Note : In such type of questions ,it is better to write `sec^(2)theta-tan^(2)thetaorcosec^(2)theta-cot^(2)theta`for1 in only numberator . If in R.H.S. the single term in either number numerator of denominator is `sin theta` then convert the question in cosec `theta and cotthetaif` single term is `costheta` then convert the question in `secthetaand tantheta`. As in this question in R.H.S . single term `sintheta` is in numerator so we will use `"cosec"^(2)theta-cot^(2)theta`for 1. L.H.S.`=(sectheta+1-tantheta)/(tantheta+1-sectheta)=(1/costheta+1-(sintheta)/(costheta))/(sintheta/(costheta)+1-(1)/(costheta))` `=(1+costheta-sintheta)/(sintheta+costheta-1)=((1)/(costheta)+(costheta)/(sintheta)-(sintheta)/(sinthe))/(sintheta/(sintheta)+(costheta)/(sintheta)-(1)/(sintheta))` (dividing Nr . and Dr . by `sintheta`to convert it in `cosecthetaandcottheta`) `=("cosec"theta+cottheta-1)/(1+cottheta-"cosec"theta)=("cosec"theta+cottheta-("cosec"^(2)theta-cot^(2)theta))/(1+cottheta-"cosec"theta)` `=(("cosec"theta+cottheta)-("cosec"theta+cottheta)("cosec"theta-cottheta))/(1+cottheta-"cosec"theta)` `=(("cosec"theta+cottheta)[1-"cosec"theta+cottheta])/(1+cottheta-"cosec"theta)="cosec"theta+cottheta` `=(1)/(sintheta)+(costheta)/(sintheta)=((1+costheta))/(sintheta)=((1+costheta)(1-costheta))/(sintheta(1-costheta))` `=(1-cos^(2)theta)/(sintheta(1-costheta))=(sin^(2)theta)/(sintheta(1-costheta))=(sintheta)/(1-cos)=R.H.S.` |
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| 30. |
`(tanA)/(1+secA) - (tanA)/(1-secA) = 2cosecA` |
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Answer» LHS =`(tanA)/(1+secA) - (tanA)/(1-secA)= (tanA(1-secA - 1 -secA))/((1+secA)(1-secA))` `=(tanA(-2secA))/(1-sec^(2)A)=(2tan A. secA)/(sec^(2)A-1)` `[therefore (a+b)(a-b)=a^(2)-b^(2)]` `=(2tanA. secA)/(tan^(2)A)` `[therefore sec^(2)A- tan^(2)A=1]` `[therefore sectheta=1/costheta` and `tantheta=(sintheta)/(costheta)]` `=(2secA)/(tanA) = 2/(sinA) = 2 cosec A= RHS` `[therefore cosectheta= 1/(sintheta)]` |
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| 31. |
If `costheta +sin theta= sqrt(2)cos theta`, then prove that` costheta-sintheta =sqrt(2)sin theta` |
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Answer» Given that , `costheta+sintheta=sqrt(2)costheta` ` rArrsintheta=sqrt(2)costheta-costheta` `rArrsintheta=(sqrt(2)-1)costheta` Multiply both sides by `(sqrt(2)+1)` `(sqrt(2)+1)sintheta=(sqrt(2)+1)(sqrt(2)-1)costheta` `rArrsqrt(2)sintheta+sintheta=costheta` `sqrt(2)sintheta=costheta-sintheta` Alternative Method : We have, `costheta+sintheta=sqrt(2)costheta` Squaring both sides , we get `(costheta+sintheta)^(2)=(sqrt2costheta)^(2)` `rArrcos^(2)theta+sin^(2)theta+2costhetasintheta=2cos^(2)theta` `rArr2cos^(2)theta-cos^(2)theta-sin^(2)theta=2sinthetacostheta` `rArr cos^(2)theta-sin^(2)theta=2sinthetacostheta` `rArr(costheta-sintheta)(costheta+sintheta)=2sinthetacostheta` `rArr(costheta-sintheta)*sqrt(2)costheta=2sinthetacostheta` `:. costheta-sintheta=sqrt(2)sintheta` |
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| 32. |
If `tantheta=(4)/(3)`, then find the value of `(3sintheta-2costheta)/(3sintheta+5costheta)` |
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Answer» `(3sintheta-2costheta)/(3sintheta+5costheta)= (3(sintheta)/(costheta)-2 (costheta)/(costheta))/(3(sintheta)/(costheta)+5(costheta)/(costheta))` (divedenumerator and denominator by `costheta`) `=(3tantheta-2)/(3tantheta+5)=(3xx(4)/(3)-2)/(3xx(4)/(3)+5)=(4-2)/(4+5)=(2)/(9)` |
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| 33. |
Prove that : `(1+costheta- sin^(2)theta)/(sintheta+sinthetacostheta)=cottheta` |
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Answer» L.H.S. `=(1+costheta-sin^(2)theta)/(sintheta+sinthetacostheta)=((1-sin^(2)theta)+costheta)/(sin theta(1+costheta))=(cos^(2)theta+costheta)/(sintheta(1+costheta))` `=(costheta(costheta+1))/(sintheta(1+costheta))=(costheta)/(sintheta)=cotthetaR.H.S.` |
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| 34. |
`(cosecA - sinA)(secA - cosA)sec^2A = tanA` |
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Answer» `L.H.S. = (cosecA-sinA)(secA-cosA)sec^2A` `=(1/sinA-sinA)(1/cosA-cosA)1/cos^2A` `=((1-sin^2A)/sinA)((1-cos^2A)/cosA)1/cos^2A` `=((cos^2A)/sinA)((sin^2A)/cosA)1/cos^2A` `=sinA/cosA` `=tanA=R.H.S.` |
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| 35. |
Porve that : (cosecA-sinA)(secA-cosA)`=(1)/(tanA+cotA)` |
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Answer» L.H.S.`=("cosec"A-sinA)(secA-cosA)` `=((1)/(sinA)-sinA/(1))((1)/(cosA)-(cosA)/(1))=((1-sin^(2)A)/(sinA))((1-cos^(2)A)/(cosA))` `=((cos^(2)A)/(sinA))((sin^(2)A)/(cosA))=cosA sinA` andR.H.S.`=(1)/(tanA+cotA)=(1)/(sinA/(cosA)+cosA/sinA)` `=(1)/((sin^(2)A+cos^(2)A)/(sinAcosA))=(sinAcosA)/(1)=sinAcosA` `:.` L.H.S.=R.H.S. |
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| 36. |
if `sinA=1/2`, then the value of `cot A`A. `sqrt(3)`B. `1/sqrt(3)`C. `sqrt(3)/2`D. 1 |
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Answer» Correct Answer - A Given, sinA=`1/2` `therefore cosA=sqrt(1-sin^(2)A)=sqrt(1-(1/2)^(2))` `=sqrt(1-1/4) = sqrt(3/4)=sqrt(3)/2` `[therefore sin^(2)A + cos^(2)=1 rArr cosA = sqrt(1-sin^(2)A)]` Now, cotA = `(cosA)/(sinA)= (sqrt(3)/2)/(1/2)= sqrt(3)` Hence, the required value of cot A is `sqrt(3)`. |
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| 37. |
If `cos A=4/5`, then the value of `tanA` isA. `3/5`B. `3/4`C. `4/3`D. `5/3` |
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Answer» Correct Answer - B Given, cosA=`4/5` `therefore sinA = sqrt(1-cos^(2)A)` `[therefore sin^(2)A+ cos^(2)A=1]` `[therefore sinA=sqrt(1-cos^(2)A)]` `=sqrt(1-(4/5)^(2))=sqrt(1-(16/25))=sqrt(9/25)=3/5` Now, `tanA=(sinA)/(cosA) = (3/5)/(4/5)=3/4` Hence, the required value of tan A is `3/4` |
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| 38. |
` (cos 70^@)/(sin 20^@) + (cos 59^@)/(sin 31^@) = 8 sin^2 30^@` |
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Answer» Here, we will use `sin(90-theta) = costheta and cos(90-theta) = sintheta` ` L.H.S. = (cos70^@/sin20^@) + (cos59^@/sin31^@)` `= (cos(90-20)^@/sin20^@) + (cos(90-31)^@/sin31^@)` `= (sin20^@/sin20^@) + (sin31^@/sin31^@)` `=1+1 = 2` `R.H.S. = 8sin^2 30^@ = 8(1/2)^2 = 2` `:. L.H.S = R.H.S.` |
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| 39. |
If `angleA` and `angleP` are acute angle such that `tanA=tanP`, then show that `angleA=angleP`. |
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Answer» Let we create a right angle triangle `ABP` with the given details. Please refer to video to see the triangle. Then, `tanA = (BP)/(AB)` `tanP = (AB)/(BP)` It is given, `tanA = tanP` `:. (BP)/(AB) = (AB)/(BP)` `=>BP^2 = AB^2` `=>BP = AB` We know, opposite angles of equal sides are also equal. `:. /_A = /_P` |
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| 40. |
If `tan theta +sin theta =m` and `tan theta -sin theta =n` then prove `m^2-n^2=4sqrt(mn)` |
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Answer» `m^2 = tan^2 theta + sin^2 theta + 2 tan theta sin theta` `n^2 = tan^2 theta + sin^2 theta - 2 tan theta sin theta` so, `m^2 - n^2 = 4 tan theta sin theta` `4 sqrt(mn) = 4sqrt((tan theta + sin theta)(tan theta - sin theta))` `= 4 sqrt(tan^2 theta - sin^2 theta)` `= 4 sqrt(sin^2 theta/cos^2 theta - sin^2 theta ` `= 4 sin theta sqrt( sec^2 theta - 1) ` `4 sqrt(mn) = 4 sin theta tan theta ` `= m^2 - n^2 ` Answer |
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| 41. |
Prove the following identity, where the angles involved are acute angles for which the expressions are defined.(iii) `(tantheta)/(1-cottheta)+(cottheta)/(1-tantheta)=1+sectheta cosectheta` |
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Answer» Left hand side =`(tantheta)/(1-cottheta)+(cottheta)/(1-tantheta)` =`(tantheta)/(1-cottheta)+(cottheta)/(1-tantheta)` =`(-tantheta)/(1-tantheta)+(cottheta)/(1-tantheta)` =`(-tan^2theta+tantheta)/(1-tantheta)` =`1/tantheta((1-tan^3theta)/(1-tantheta))` =`1/tantheta(1+tan^2theta+tantheta)` =`1+seccthetacosectheta` |
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| 42. |
`(cos45)/(sec30+cosec30)` |
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Answer» `(cos45^(@))/(sec30^(@)+"cosec"30^(@))=(cos45^(@))/(sec30^(@)+"cosec"30^(@))=(1/(sqrt(2)))/(2/sqrt(3)+2)=(1/(sqrt(2)))/(((2+2sqrt(3)))/(sqrt(3)))` `=(1)/(sqrt(2))xx(sqrt(3))/(2(sqrt(3)+1))= (sqrt(3)(sqrt(3)-1))/(2sqrt(2)(sqrt(3)+1)(sqrt(3)-1))` `=(sqrt(6)(sqrt(3)-1))/(4(3-1))=(sqrt(6)(sqrt(3)-1))/(8)` |
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| 43. |
Find the vlaue of `theta` if `2cos3theta=1and0^(@)ltthetalt90^(@)`. |
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Answer» `2cos3theta=1` `rArrcos3theta=(1)/(2)` `rArr cos3theta=cos60^(@)` `rArr3theta=60^(@)` `rArrtheta=20^(@)` |
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| 44. |
If `2sin^(2)theta-cos^(2)theta=2`, then find the value of `theta`. |
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Answer» `2sin^(2)theta=cos^(2)theta=2` `rArr2sin^(2)theta-(1-sin^(2)theta)=2` `rArr2sin^(2)theta-1+sin^(2)theta=2` `rArr3sin^(2)theta=3rArrsin^(2)theta=1` `rArrsin theta=1rArrtheta=90^(@)` |
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| 45. |
Solve : `2sin^(2)theta=(1)/(2),0^(@)ltthetalt90^(@)`. |
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Answer» `2sin^(2)theta=(1)/(2)` `rArrsin^(2)theta=(1)/(2xx2)=(1)/(4)` `rArrsintheta=sqrt((1)/(4))=(1)/(2)` (taking positive sign only) `rArrsinthetasin30^(@)` `rArrtheta=30^(@)` |
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| 46. |
If `sinx+sin^2x+sin^3x=1` then find the value of `cos^6x-4cos^4x+8cos^2x` |
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Answer» We have , `sinx+sin^(2)x+sin^(3)x=1` `rArrsinx+sin^(3)x=1-sin^(2)x` `rArrsinx(1+sin^(2)x)=cos^(2)x` `rArrsinx(1+1-cos^(2)x)=cos^(2)x` `rArrsinx(2-cos^(2)x)=cos^(2)x` Squaring both sides ,we get `sin^(2)x(2-cos^(2)x)^(2)=cos^(4)x` `rArr(1-cos^(2)x)(4+cos^(4)x-4cos^(2)x)=cos^(4)x` `rArr4+cos^(4)x-4cos^(2)x-4cos^(2)x-cos^(6)x+4cos^(4)x=cos^(4)x` `rArrcos^(6)x-4cos^(4)x+8cos^(2)x-4=0` |
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| 47. |
Prove that `cos^2 theta cosec^2 theta+sin^2 theta=cosec^@ theta` |
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Answer» L.H.S. `=cos^(2)theta*cosectheta +sin theta = cos^(2)theta*(1)/(sintheta)+sintheta` `=(cos^(2)theta)/(sintheta)+(sintheta)/(1)=(cos^(2)theta+sin^(2)theta)/(sintheta)=(1)/(sintheta)` `=cosectheta = R.H.S.` |
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| 48. |
Smlplify : `(1+tan^(2)theta)(1-sintheta)(1+sintheta)` |
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Answer» `(1+tan^(2)theta)(1-sintheta)(1+sintheta)` `=(1+tan^(2)theta)[(1)^(2)-(sintheta)^(2)]=(1+tan^(2)theta)(1-sin^(2)theta)` `=sec^(2)theta*cos^(2)theta` (using the identities `sec^(2)theta=1+tan^(2)thetaandsin^(2)theta+cos^(2)theta=1)` `(1)/(cos^(2)theta)*cos^(2)theta=1` |
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| 49. |
Find the value of `theta` if `sec^(2)theta+tan^(2)theta=(5)/(3)`and`theta` lies in first quadrant. |
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Answer» `sec^(2)theta+tan^(2)theta=(5)/(3)` `rArr1+tan^(2)theta+tan^(2)theta=(5)/(3)rArr2tan^(2)theta=(5)/(3)-1=(2)/(3)` `rArrtan^(2)theta=(1)/(3)rArrtan^(2)theta=((1)/(sqrt(3)))^(2)` `rArrtantheta=(1)/(sqrt(3))` (taking positive sign only) `rArrtantheta=tan30^(@)rArrtheta=30^(@)` |
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| 50. |
Prove that : `sqrt((1-cos^(2)theta)sec^(2)theta)=tantheta` |
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Answer» L.H.S. `=sqrt((1-cos^(2)theta)sec^(2)theta)=sqrt(sin^(2)theta*(1)/(cos^(2)theta))=sqrt(sin^(2)theta/(cos^(2)theta))` `=sqrt(tan^(2)theta)=tan theta=R.H.S.` |
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