InterviewSolution
Saved Bookmarks
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=90^(@)` then prove that `1+(tanA)/(tanB)=sec^(2)A.` |
|
Answer» `A+B=90^(@)impliesB=90^(@)-A` `:.1+(tanA)/(tanB)=1+(tanA)/(tan(90^(@)-A))` `=1+(tanA)/(cotA)=1+tanA.tanA` `=1+tan^(2)A=sec^(2)A`. Hence `1+(tanA)/(tanB)=sec^(2)A.` |
|
| 2. |
If `A+B=90^(@)andtanA=(3)/(4),` then value of cot B isA. `(3)/(4)`B. `(4)/(3)`C. `(3)/(5)`D. `(4)/(5)` |
|
Answer» Correct Answer - A Given that `A+B=90^(@)or,A=90^(@)-B` `:.tanA=(3)/(4)impliestan(90^(@)-B)=(3)/(4)impliescot=(3)/(4)` |
|
| 3. |
If `A+B=90^(@)`, then the value of `sin^(2)A+sin^(2)B=`A. `-1`B. 0C. 1D. `(1)/(sqrt2)` |
|
Answer» Correct Answer - C `A+B=90^(@)impliesB=90^(@)-A` Now, `sin^(2)+A+sin^(2)B` `=sin^(2)A+cos^(2)A=1` |
|
| 4. |
If A and B are acute angles such that `sin A =cos B`, prove that `(A+B)=90^@`. |
|
Answer» `sin A=cos B rArr sin A =sin (90^@-B)` `rArr A=90^@-B [because A and (90^@-B) "are acute"]` `rArr A+B=90^@`. |
|
| 5. |
AOB is a diameter of a circle with centre O and C is any point on the circle, joining A, C, B, C, and O, C, prove that `cosec^(2)angleCAB-1=tan^(2)angleABC`. |
|
Answer» since AOB is a diameter and C is any point on the circle, `:.angleACB` is a semicircular angle. `:.angleACB=90^(@)` `:.` AB is a hypotenuse of the right-angled triangle ABC. Again, since `angleACB=90^(@),:.angleBAC+angleCBA=90^(@)`. `cosec^(2)angleCAB-1` `=cot^(2)angleCAB` `=cot^(2)(90^(@)-angleABC)` `=tan^(2)angleABC`. Hence `cosec^(2)angleCAB-1=tan^(2)angleABC`. |
|
| 6. |
Find the value of `sin^(2)5^(@)+sin^(2)10^(@)+sin^(2)15^(@)+.............+sin^(2)90^(@)`. |
|
Answer» `sin^(2)5^(@)+sin^(2)10^(@)sin^(2)15^(@)+.............sin^(2)90^(@)`. `=(sin^(2)5^(@)sin^(2)85^(@))+(sin^(2)10^(@)+sin^(2)80^(@))+(sin^(2)15^(@)+sin^(2)75^(@))+(sin^(2)20^(@)+sin^(2)70^(@))+(sin^(2)25^(@)+sin^(2)65^(@))+(sin^(2)30^(@)+sin^(2)60^(@))+` `(sin^(2)35^(@)+sin^(2)55^(@))+(sin^(2)40^(@)+sin^(2)50^(@))+sin^(2)45^(@)+sin^(2)90^(@)` `={sin^(2)5^(@)+sin^(2)(90^(@)-15^(@))}+...................sin^(2)45^(@)+sin^(2)90^(@)` `=(sin^(2)5^(@)+cos^(2)5^(@))+(sin^(2)10^(@)+cos^(2)15^(@))+...................+sin^(2)45^(@)+sin^(2)90^(@)` `=1+1+1+...............` (upto 8 terms) `+((1)/(sqrt(2)))^(2)+1` `=8+(1)/(2)+1=9(1)/(2)`. Hence the required value `=9(1)/(2)` |
|
| 7. |
If `sec 5A=cosec (A-30^@)`, where 5A is an acute angle then find the value of A. |
|
Answer» `sec 5A=cosec (A-30^@)` `rArr cosec (90^@-5A)=cosec(A-30^@) [because sec theta =cosec (90^@-theta)]` `rArr 90^@-5A =A-30^@rArr 6A=120^@rArr A =20^@`. Hence, `A=20^@`. |
|
| 8. |
If`s in3A" "=" "cos(A 26o)`,where 3A is an acute angle, find the value of A. |
|
Answer» Correct Answer - `A=29^@` `sin 3A=cos(A-26^@)rArr (90^@-3A)=cos (A-26^@)rArr 90^@-3A=A-26^@`. |
|
| 9. |
If `sin (theta+34^@)=cos theta and (theta +34^@)` is acute show that `theta =28^@`. |
|
Answer» `sin (theta +34^@)=cos theta` `rArr sin (theta +34^@)=sin (90^@-theta) [because cos theta =sin (90^@-theta)]`. `rArr theta + 34^@=90^@-theta rArr 2 theta =(90^@-34^@)=56rArr theta =28^@`. Hence,`theta =28^@`. |
|
| 10. |
If `sin 3A =cos(A-10^(@))` where 3A is an acute angle , then find the value of A. |
|
Answer» `sin 3A=cos (A-10^@)` `rArr (90^@-3A)=cos(A-10^@)[because sin theta =cos (90^@0theta)]` ` rArr 90^@-3A =A -10^@rArr 4A=100^@rArr A=25^@`. Hence, `A=25^@`. |
|
| 11. |
Without using trigonometric tables, evaluate each of the following:`(cos^2 20^0+cos^2 70^0)/(sec^2 50^0-cot^2 40^0)+2cos e c^2 58^0-2cot58^0tan32^0-4tan13^0tan37^0tan45^0tan53^0tan77^0` |
|
Answer» Given expression `(cos ^2 20+cos^2 70^@)/(sec^2 50^@-cot ^2 40^@)+2cosec ^2 58^@-2cot58^@tan 32^@-4tan 13^@tan 37^@tan 45^@tan 53^@tan 77^@`. ` =(cos^2 20^@+[cos (90^@-20^@)]^2)/(sec^2 50^@-[cot(90^@-50^@)]^2)` `+2[cosec ^2 58^@-cot 58^@tan(90^@-58^@)]` ` -4 (tan 13^@tan 77^@)(tan 37^@tan 53^@)tan 45^@` `=(cos^2 20^@+sin^2 20^@)/(sec^2 50^@-tan^2 50^@)+2[cosec^2 58^@-cot^2 58^@]` ` -4tan 13^@tan(90^@-13^@).tan 37^@tan (90^@-37^@).tan 45^@ [because cos (90^@-theta)=sin theta, cot (90^@-theta) = tan theta and tan (90^@-theta )=cot theta ]` ` =(1)/(1) +(2xx 1) -4(tan 13^@cot 13^@).(tan 37^@ cot 37^@).1` `[because cos^2 theta +sin^2 theta -tan^2 theta=1, cosec^2 theta -cot^2 theta =1]` `=(3-4xx1xx1xx1)=(3-4)=-1`. |
|
| 12. |
The value of `(sin43^(@)cos47^(@)+cos43^(@)sin47^(@))` is |
|
Answer» Correct Answer - B `sin43^(@)cos47^(@)+cos43^(@)sin47^(@)` `=sin43^(@)cos(09^(@)-43^(@))+cos43^(@)sin(90^(@)-43^(@))` `=sin43^(@)sin43^(@)+cos43^(@)cos43^(@)` `=sin^(2)43^(@)+cos^(2)43^(@)=1`. |
|
| 13. |
Evaluate : `(2sin^2 63^@+1+2sin^2 27^@)/(3 cos ^2 17^@-2+3cos^2 73^@)` |
|
Answer» Given exrpession `(2sin^2 63^@+1+2sin^2 27^@)/(3 cos ^2 17^@-2+3cos^2 73^@)=(2(sin^2 63^@+sin^2 27^@)+1)/(3(cos^2 17^@+cos^2 73^@)-2)` `=(2[sin^2 63^@+sin^2 (90^@-63^@)]+1)/(3[cos^2 17^@+cos^2 (90^@-17^@)]-2)` `=(2[sin^2 63^@+cos^2 63^@]+1)/(3[cos^2 17^@+sin^2 17^@]-2)` `[because sin (90^@-theta)=cos theta and cos (90^@-theta )=sin theta ]` `=((2xx1)+1)/((3xx1)-2)=(2+1)/(3-2)=(3)/(1)=3`. |
|
| 14. |
Find the value of `(2sin^(2)63^(@)+1+2sin^(2)27)/(3cos^(2)17^(@)-2+3cos^(2)73^(@))` |
|
Answer» We have, `(2sin^(2)63^(@)+1+2sin^(2)27)/(3cos^(2)17^(@)-2+3cos^(2)73^(@))` `=(2sin^(2)63^(@)+2sin^(2)(90^(@)-63^(@))+1)/(3cos^(2)17^(@)-2+3cos^(2)(90^(@)-17^(@)))=(2sin^(2)63^(@)+2cos^(2)63^(@)+1)/(3cos^(2)17^(@)-2+3sin^(2)17^(@))` `=(2(sin^(2)63^(@)+cos^(2)63^(@))+1)/(3(cos^(2)17^(@)+sin^(2)+17^(@))-2)=(2xx1+1)/(3xx1-2)` `=(2+1)/(3-2)=(3)/(1)=3`. Hence the required value is 3. |
|
| 15. |
`sec70^(@)sin20^(@)+cos20^(@)cosec70^(@)=2` |
|
Answer» `sec70^(@)sin(90^(@)-70^(@))+cos20^(@)cosec(90^(@)-20^(@))` `=sec70^(@)cos70^(@)+cos20^(@)sec20^(@)` `=(1)/(cos70^(@)).cos70^(@)+cos20^(@).(1)/(cos20^(@))` =1+1=2. Hence `sec70^(@)sin20^(@)+cos20^(@)cosec70^(@)=2`. |
|
| 16. |
If `sin51^(@)=(a)/sqrt(a^(2)+b^(2))` then find the value of `tan39^(@)` |
|
Answer» `sin51^(@)=(a)/sqrt(a^(2)+b^(2))orsin^(2)51^(@)=(a^(2))/(a^(2)+b^(2))` or, `sin^(2)(90^(@)-39^(@))=(a^(2))/(a^(2)+b^(2))or,cos^(2)39^(@)=(a^(2))/(a^(2)+b^(2))` or, `sec^(2)39^(@)(a^(2)+b^(2))/(a^(a))or,1+tan^(2)39^(@)=(a^(2)+b^(2))/(a^(2))` `tan^(2)39^(@)=(a^(2)+b^(2))/(a^(2))-1` `tan^(2)39^(@)=(a^(2)+b^(2)-a^(2))/(a^(2))ortan^(2)39^(@)=(b^(2))/(a^(2))` or, `tan39^(@)=(b)/(a)` Hence `tan39^(@)=(b)/(a)`. |
|
| 17. |
If `sin17^(@)=(x)/(y)` then prove that `sec17^(@)-sin73^(@)=(x^(2))/(ysqrt(y^(2)-x^(2)))` |
|
Answer» `LHS=sec17^(@)-sin73^(@)` `(1)/(cos17^(@))-sin(90^(@)-17^(@))` `=(1)/(cos17^(@))-cos17^(@)=(1-cos^(2)17^(@))/(cos17^(@))` `=(sin^(2)17^(@))/(sqrt(1-sin^(2)17^(@)))=(((x)/(y))^(2))/(sqrt(1-((x)/(y))^(2)))` Hence `sec17^(@)-sin73^(@)=(x^(2))/(ysqrt(y^(2)-x^(2)))` |
|
| 18. |
Prove that `cot12^(@)cot38^(@)cot52^(@)cot78^(@)cot60^(@)=(1)/(sqrt3)` |
|
Answer» `cot12^(@)cot38^(@)cot52^(@)cot78^(@)cot60^(@)` `=(cot12^(@)cot78^(@))(cot38^(@)cot52^(@))cot60^(@)` `={cot12^(@)cot(90^(@)-12^(@))}{cot38^(@)cot(90^(@)-38^(@))}cot60^(@)` `=(cot12^(@)tan12^(@))(cot38^(@)tan38^(@))cot60^(@)`. `=1xx1xx(1)/(sqrt3)=(1)/(sqrt3)` Hence `cot12^(@)cot38^(@)cot52^(@)cot78^(@)cot60^(@)=(1)/(sqrt3)` |
|
| 19. |
`tan 1^@ tan2^@...tan 89^@=` |
|
Answer» We have `tan 1^@ tan 2^@ tan 3^@.....tan 89^@` `=tan 1^@tan 2^@.....tan44^@tan 45^@tan 46^@.....tan tan 88^@tan 89^@` ` =(tan 1^@tan 89^@)(tan 2^@.tan 2^@.tan 88^@).....(tan 44^@.tan 46^@).tan 45^@` `={tan 1^@.tan (90^@-1^@)}.{tan 2^@.tan (90^@-2)}.....{tan 44^@.tan (90^@-44^@)}.tan 45^@` `=tan 1^@.cot 1^@)(tan 2^@.cot 2^@)......(tan 44^@cot 44^@).1 [because tan (90^@-theta) =cot theta and tan 45^@=1]` ` =1xx1 xx....xx 1 xx 1=1`. |
|
| 20. |
Prove that `cosec^(2)22^(@)cot^(2)68^(@)=sin^(2)22^(@)+sin^(2)68^(@)+cot^(2)68^(@)`. |
|
Answer» `cosec^(2)22^(@)cot^(2)68^(@)=cosec^(2)(90^(@)-68^(@))cot^(2)68^(@)` `=sec^(2)68^(@)cot^(2)68^(@)=(1+tan^(2)68^(@))cot^(2)68^(@)` `=cot^(2)68^(@)+tan^(2)68^(@)cot^(2)68^(@)` `=cot^(2)68^(@)+tan^(2)68^(@)cot^(2)68^(@)` `=cot^(2)68^(@)+tan^(2)68^(@).(1)/(tan^(2)68^(@))` `=cot^(2)68^(@)+1` `=cot^(2)68^(@)+sin^(2)22+cos^(2)(90^(@)-68^(@))` `=cot^(2)68^(@)+sin^(2)22+sin^(2)68^(@)` `=sin^(2)22^(@)+sin^(2)68^(@)+cot^(2)68^(@)`. Hence `cosec^(2)22^(@)cot^(2)68^(@)=sin^(2)22^(@)+sin^(2)68^(@)+cot^(2)68^(@)` |
|
| 21. |
Evaluate: ` (cos 58^@)/(sin 32^@)+(sin 22^@)/(cos 68^@)-(cos 38^@cosec 52^@)/(tan 18^@tan 35 ^@ tan 60^@tan 72^@tan 55^@)`. |
|
Answer» Given expression ` (cos 58^@)/(sin 32^@)+(sin 22^@)/(cos 68^@)-(cos 38^@cosec 52^@)/(tan 18^@tan 35 ^@ tan 60^@tan 72^@tan 55^@)`. ` =(cos 58^@)/(sin(90^@-58^@))+(sin 22^@)/(cos (90^@-22^@))-(cos 38^@cosec(90^@-38^@))/((tan 18^@tan 72^@)(tan 35^@tan 55^@)tan 60^@)` ` =(cos 58^@)/(cos 58^@)+(sin 22^@)/(sin 22^@)` ` -(cos 38^@sec 38^@)/({tan 18^@tan(90^@-18^@)}{tan 35^@-tan(90^@-35^@)}tan 60^@)` `=(1+1)-(1)/((tan 18^@cot 18^@)(tan 35^@cot 35^@)sqrt(3))` ` =(2-(1)/(sqrt(3)))=(2sqrt(3)-1)/(sqrt(3))=(2sqrt(3)-1)/(sqrt(3))xx (sqrt(3))/(sqrt(3))=(1)/(3)(6-sqrt(3))`. |
|
| 22. |
`(sin52^(@)+cos38^(@))/(sin38^(@)+cos52^(@))`A. `cot52^(@)`B. `sin52^(@)`C. `cossec52^(@)`D. `tan52^(@)` |
| Answer» Correct Answer - d | |
| 23. |
Find the value of `cot17^(@)(cot73^(@)cos^(2)22^(@)+(1)/(tan73^(@)sec^(2)68^(@)))` |
|
Answer» `cot17^(@)(cot73^(@)cos^(2)22^(@)+(1)/(tan73^(@)sec^(2)68^(@)))` `=cot17^(@)(cot73^(@)cos^(2)22^(@)+cot73^(@)cos^(2)68^(@))` `=cot(90^(@)-73^(@)){cot73^(@)cos^(2)22^(@)+cot73^(@)cos^(2)(90^(@)-22^(@))}` `=tan73^(@).cot73^(@)(cos^(2)22^(@)+sin^(2)22^(@))` `=tan73^(@).(1)/(tan73^(@))xx1=1xx1=1`. Hence the required value =1. |
|
| 24. |
The value of `cos54^(@)andsin36^(@)` are equal. |
| Answer» since `cos54^(@)=cos(90^(@)-36^(@))=sin36^(@)` | |
| 25. |
The values of `sin72^(@)andcos108^(@)` are equal. |
| Answer» since `sin72^(@)=sin(180^(0)-108^(@))=sin108^(@)` | |
| 26. |
The simplified value of `(sin12^(@)-cos78^(@))` is 1. |
|
Answer» since `(sin12^(@)-cos78^(@))` `=sin12^(@)-cos(90^(@)-12^(@))` `=sin12^(@)-sin12^(@)=0` |
|
| 27. |
The value of `(sin12^(@)xxcos18^(@)xxsec78^(@)xxcosec72^(@))` is _______ |
|
Answer» We have, `sin12^(@)xxcos18^(@)xxsec78^(@)xxcosec72^(@)` `=sin12^(@)xxcos18^(@)xxsec(90^(@)-12^(@))xxcosec(90^(@)xx18^(@))` `=sin12^(@)xxcos18^(@)xx(1)/(sin12^(@))=1xx1=1`. |
|
| 28. |
If A and B are complementary to each other, then sin A ______ |
|
Answer» cos B , since A and B are complementary, then `A+B=90^(@)or,A=90^(@)-B` `impliessinA=sin(90^(@)-B)=cosB` |
|
| 29. |
`cos72^(@)-sin18^(@)=` ________ |
|
Answer» 0 , since `cos72^(@)-sin18^(@)` `=cos72^(@)-sin(90^(@)-72^(@))` `=cos72^(@)-cos72^(@)=0` |
|
| 30. |
If `sin10theta=cos8thetaand10theta` is a positive acute angle, then find the value of `tan9theta`. |
|
Answer» Given that `sin10theta=cos8thetaand10theta` is acute. `:.sin10theta=sin(90^(@)-8theta)` `implies10theta=90^(@)-8theta` `implies10theta+8theta=90^(@)` `implies18theta=90^(@)` `impliestheta=(90^(@))/(18)` `impliestheta=5^(@)` `implies9theta=9xx5^(@)=45^(@)` `impliestan9theta=tan45^(@)` `impliestan9theta=1` Hence the value of `tan9theta=1` |
|
| 31. |
Expresss each of the following in terms of T-ratios of angles lying between `0^theta and 45^@`. (i) `sin 67^@+cos 75^@` (ii) ` cot 65^@+tan 49^@` (iii) ` sec 78^@+cosec 56^@` (iv) `cosec 54^@+sin 72^@`. |
|
Answer» Correct Answer - `(i) cos 23^@+sin 15^@` `(ii) tan 25^@+cot 41^@` `(iii) cosec 12^@+sec34^@` `(iv) sec 36^@+cos 18^@` `(i) sin 67^@+cos 75^@=sin (90^@-23^@)+cos (90^@-15^@)=cos23^@+sin 15^@`. `(ii) cot 65^@+tan 49^@=cot (90^@-25^@)+tan (90^@-41^@)=tan 25^@+cot 41^@`. |
|
| 32. |
If A, B, C are the angles of a triangle then `tan ((B+C)/2)=`A. `sin""(A)/(2)`B. `cot""(A)/(2)`C. `cos""(A)/(2)`D. `sec""(A)/(2)` |
| Answer» Correct Answer - b | |