Explore topic-wise InterviewSolutions in .

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.

151.

If `alphaa n dbeta`are the solutions of the equation `at a ntheta+bs e ctheta=c ,`then show that `tan(alpha+beta)=(2a c)/(a^2-c^2)`

Answer» `atantheta+bsectheta=c`
`rArr bsectheta=c-a tan theta`
`rArr b^(2)sectheta=(c-a tantheta)^(2)`
`rArr b^(2)(1+tan^(2)theta=c^(2)+a^(2)tantheta-2ac tantheta`
`rArr tan^(2)theta(b^(2)-a^(2))+2actantheta + (b^(2)-c^(2))=0`
Now, the roots of the equation be `tanalpha` and `tanbeta`,
`therefore tanalpha + tanbeta=(-2ac)/(b^(2)-a^(2))`
and `tanalpha. tanbeta=(tanalpha+tanbeta)/(1-tanalphatanbeta)`
`=((-2ac)/(b^(2)-a^(2)))/(1-(b^(2)-c^(2))/(b^(2)-a^(2)))=(-2ac)/((b^(2)-a^(2))-(b^(2)-c^(2)))`
Discussion
152.

Find the value of `sin37(1/2)^(@).sin7 (1/2)^(@)`

Answer» `sin37 (1/2)^(@). Sin7(1/2)^(@)`
`=1/2.2sin37 (1/2)^(@).sin 7(1/2)^(@)`
`=1/2.[cos37(1/2)^(@)-7(1/2)^(@)-cos37(1/2)^(@)+7(1/2)^(@)]`
`=1/2(cos30^(@)-cos45^(@))`
`=1/2(sqrt(3)/2-1/sqrt(2))=(sqrt(3)-sqrt(2))/(4)` Ans.
Discussion
153.

In `DeltaABC`, prove that: `asin(B-C)+bsin(C-A)+csinA-B)=0`

Answer» 1st term =`asin(B-C)`
`=ksinA.sin(B-C)`
`=ksin{180^(@)-(B+C)}sin(B-C)`
`ksin(B+C).sin(B-C)`
`=ksin^(2)B-sin^(2)C`
Similarly,
2nd term `=k(sin^(2)C-sin^(2)A)`
3rd term `=k(sin^(2)A-sin^(2)B)`
Now L.H.S. `=asin(B-C)+bsin(C-a)+csin(A-B)`
`=k(sin^(2)B-sin^(2)C)+k(sin^(2)C-sin^(2)A)+k(sin^(2)A-sin^(2)B)`
`ksin^(2)B-ksin^(2)C-ksin^(2)A+ksin^(2)A-ksin^(2)B` =0 = RHS Hence Proved.
Discussion
154.

If a,b,c are in A.P, then prove that: `2sinA/2.sinC/2=sinB/2`

Answer» Let `2sinA/2.sinC/2=sinB/2`
`rArr 2sqrt((s-b)(s-c))/(bc).sqrt((s-a)(s-b))/(ab)=sqrt((s-a)(s-c))/(ac)`
`rArr 2(s-b)/b.sqrt((s-a)(s-c))/(ac)=sqrt((s-a)(s-c))/(ac)`
`rArr 2(s-b)/(b)=1`
`rArr 2s-2b=b`
`rArr a+b+c-2b=b`
`rArr a+c=2b`
`rArr a+c=2b`
`rArr` a,b,c are in A.P.
Which is given.
Therefore, `2sinA/2.sinC/2=sinB/2`, if a,b,c are in A.P. Hence proved.
Discussion
155.

Prove that: `(sin(A-C)+2sinA+sin(A+C))/(sin(B-C)+2sinB+sin(B+C))=(sinA)/(sinB)`

Answer» LHS =`(sin(A-C)+2sinA+sin(A+C))/(sin(B-C)+2sinB+sin(B+C))`
`=(sin(A+C)+sin(A-C)+2sinA)/(sin(B+C)+sin(B-C)+2sinB)`
`(2sin(A+C+A-C)/(2).cos(A+C-A+C)/(2)+2sinA)/(2sin((B+C+B-C)/2).cos(B+C-B+C)/(2)+2sinB)`
`=(2sinAcosC+2sinA)/(2sinBcosC+2sinB)`
`(sinA(2cosC+2))/(sinB(2cosC+2))`
`=(sinA(2cosC+2))/(sinB(2cosC+2))`
`=(sinA)/(sinB)` =RHS Hence Proved.
Discussion
156.

In `DeltaABC`, prove that: `sinA+sinB-sinC=4sinA/2sinB/2cosC/2`

Answer» LHS `=sinA+sinB-sinC`
`=2sin(A+B)/(2).cos(A-B)/2-2sinC/2cosC/2`
`=2sin(180^(@)-C)/(2)cos(A-B)/2(2)` `(therefore A+B+C=180^(@))`
`=2sin(90^(@)-C/2)cos(A-B)/(2)-2sin{90^(@)-(A+B)/(2)}cosC/2`
`=2cosC/2cos(A-B)/(2)-2cos(A+B)/(2)cosC/2`
`=2cosC/2[cos(A-B)/(2)-cos(A+B)/2]`
`=2cosC/2.2sinA/2sinB/2`
`=4sinA/2sinB/2cosC/2`=RHS Hence Proved.
Discussion
157.

Prove that: `(cos5A+cos3A+cosA)/(sin5A-sin3A+sinA)=cotA`

Answer» LHS `=(cos5A+cosA+cosA)(sin5A-sin3A+sinA)`
`=(2cos(5A+3A)/2. cos(5A-3A)/(2)+cosA)/(2cos(5A+3A)/2.cos(5A-3A)/(2)+sinA)`
`=(2cos4AcosA+cosA)/(2cos4AsinA+sinA)`
`=(cosA(2cos4A+1))/(sinA(2cos4A+1))=cotA`=RHS Hence Proved.
Discussion
158.

i) In `DeltaABC`, if b=17, c=11 and `angleA=60^(@)`, find `tan(B-C)/(2)` ii) In `DeltaABC, angleB=90^(@)`, then prove that: `tanA/2=sqrt((b-c)/(b+c))`

Answer» `(3sqrt(3))/14`
Discussion
159.

In `DeltaABC,` if b=7, c=24, a=25, find `angleA`.

Answer» `angleA=90^(@)`
Discussion
160.

In `DeltaABC, angleA=60^(2), angleB=45^(@)`, find `a:b`.

Answer» `sqrt(3):sqrt(2)`
Discussion
161.

In `DeltaABC, cotA/2, cotB/2, cotC/2` are in A.P., then the true statement is:A. `b^(2)=ac`B. `c^(2)=ab`C. `2b=a+c`D. `2a=b+c`

Answer» Correct Answer - C
Discussion
162.

In `DeltaABC, angleB=90^(@), angleA=30^(@), b=20cm`, find a and c.

Answer» a=10cm, c=`10sqrt(3)` cm
Discussion
163.

In `DeltaABC, a=2, b=sqrt(6)` and `c=sqrt(3)+1`, find `angleA`.

Answer» `cosA=(b^(2)+c^(2)-a^(2))/(2bc)`
`=(sqrt(6))^(6)+((sqrt(3)+1)^(2)-2^(2))/(2sqrt(6)(sqrt(3)+1))`
`=(2sqrt(3)(sqrt(3)+1))/(2sqrt(6)(sqrt(3)+1))`
`=1/sqrt(2)=cos45^(@)`
`A=45^(@)`.
Discussion
164.

In `DeltaABC`, `a^(2),b^(2),c^(2)` are in A.P. Prove that cotA, cotB, cotC are also in A.P.

Answer» Let cotA, cotB, cotC are in A.P.
`rArr (cosA)/(sinA),(cosB)/(sinB)` are in A.P.
`rArr (b^(2)+c^(2)-a^(2))/(2bc.a/k),(c^(2)+a^(2)-b^(2))/(2ca.b/k), (a^(2)+b^(2)-c^(2))/(2ab.c/k)` are in A.P.
`rArr b^(2)+c^(2)-a^(2),c^(2)+a^(2)-b^(2),a^(2)+b^(2)-c^(2)` are also in A.P(Multiply each term by `(2abc)/(k))`
`rArr -2a^(2),-2b^(2),-2c^(2)` are in A.P. (Subtract `a^(2)+b^(2)+c^(2)` from each term)
`rArr a^(2),b^(2),c^(2)` are in A.P. (Divide each term by `-2`)
`therefore` cotA, cotB, cotC are in A.P. Hence Proved.
Discussion
165.

In `DeltaABC`, prove that: `(b+c)cos(B+C)/(2)=acos(B-C)/(2)`

Answer» LHS `=(b+c)cos(B+C)/(2)`
`=(ksinB+ksinC).cos(pi-A)/(2)`
`k.(sinB+sinC).sinA/2`
`k.2sin(B+C)/(2).cos(B-C)/(2).sinA/2`
`k.2sin(pi-A)/(2).sinA/2.cos(B-C)/(2)`
`=k.(2cosA/2sinA/2)cos(B-C)/(2)`
`=k.sinAcos(B-C)/(2)`
`=acos(B-C)/(2)` = RHS Hence Proved.
Discussion
166.

In `DeltaABC, (b+c)/(11) = (c+a)/(12)=(a+b)/(13)`, prove that: `cosA:cosB:cosC=?`A. `25:19:17`B. `9:7:25`C. `7:9:25`D. `7:19:25`

Answer» Correct Answer - D
Discussion
167.

If the ratio of angles of `DeltaABC` is `1:2:3`, find the ratio of its sides.

Answer» Correct Answer - `1:sqrt(3):2`
Discussion
168.

Solve the equation: `sin2theta+cos3theta=0`

Answer» `theta=2npi+pi//2,theta=1/5(2npi-pi/2)`
Discussion
169.

In `DeltaABC`, prove that: `sin(A+B/2).cosB/2=(c+a)/(a+b)cosC/2.cos(A-B)/(2)`

Answer» RHS. `=(c+a)/(a+b).cosC/2.cos(A-B)/(2)`
`(ksinc+ksinA)/(ksinA+ksinB).cosC/2cos(A-B)/(2)`
`=(2sin(C+A)/(2).cos(C-A)/(2))/(2ain(A+B)/(2).cos(A-B)/(2)).cosC/2.cos(A-B)/(2)`
`=(sin(180^(@)-B)/(2).cos(180^(2)-(A+B)-A)/(2))/(sin(180^(@)-C)/(2)).cos(C/2)`
`=(sin(90^(@)-B/2)cos{90^(@)-(A+B/2)})/(sin(90^(@)-C/2)cosC/2`
`=(cosB/2.sin(A+B/2))/(cos(C/2)).cos(C/2)`
`=cos(B/2).sin(A+B/2)`=LHS Hence Proved.
Discussion
170.

In `DeltaABC, acos^(2)C/2+ccos^(2)A/2=(3b)/(2)`, then prove that the sides of `DeltaABC`, are in A.P.

Answer» `acos^(2)C/2+cos^(2)A/2=(3b)/(2)`
`rArr (a.(1+cosC))/(2)+(c.(1+cosA))/(2)=(3b)/(2)`
`rArr a(1+cosC)+c(1+cosA)=3b`
`rArr a+acosC+c+(c)cosA=3b`
`rArr a+c+(acosC+(c)cosA)=3b`
`rArr a+c+b=3b`
`rArr a+c+B=3b`
`rArr a+c=2b`
`rArr a+c=2b`
`rArr` a,b,c are in A.P. Hence Proved.
Discussion
171.

In `DeltaABC`, prove that: `a(cosC-cosB)=2(b-c)cos^(2)A/2`

Answer» LHS `=a(cosC-cosB)`
`k.sinC.2sin(C+B)/(2).sin(B-C)/(2)`
`=k.2sinA/2cosA/2.2sin(pi-A)/(2).sin(B-C)/(2)`
`=k.2sin(pi-(B+C))/(2)sin(B-C)/(2). cosA/2.2cosA/2`
`=k.(2cos(B+C)/(2).sin(B-C)/(2)).2cos^(2)A/2`
`=2k. (sinB-sinC).cos^(2)A/2`
`=2(b-c).cos^(2)A/2=`RHS Hence Proved.
Discussion
172.

Solve: `4sin^(4)theta+cos^(4)theta=1`

Answer» `theta=npi,theta=npi+-alpha`, where `sin^(2)alpha=2/5`
Discussion
173.

In `DeltaABC`, prove that: `(c-bcosA)/(b-ccosA)=(cosB)/(cosC)`

Answer» LHS `=(c-bcosA)/(b-ccosA)`
`=((acosB+bcosA)-bcosA)/((acosC+ccosA)-ccosA)` (From projection formula)
`=(acosB)/(acosC)=(cosB)/(cosC)= RHS` Hence Proved.
Discussion
174.

If `x+y+z=xyz`, prove by trignometry that: `(2x)/(1-x^(2))+(2y)/(1-y^(2))+(2z)/(1-z^(2))=(2x)/(1-x^(2)).(2y)/(1-y^(2)).(2z)/(1-z^(2))`

Answer» Let x=tanA, y=tanB, z=tanC
Now, `x+y+z=xyz`
`rArr x+y=-z+xyz`
`=-z(1-xy)`
`rArr (x+y)/(1-xy)=-z`
`rArr (tanA+tanB)/(1-tanAtanB)=-tanC`
`rArr tan(A+B)=tan(pi-C)`
`rArr 2A+2B=2pi-2C`
`rArr tan(2A+2B)=tan(2pi-2C)`
Discussion