Q.If A+B+C=0, then prove that 1cosCcosBcosC1cosAcosBcosA1=0
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Determinant Evaluation Using Identities
Determinant Evaluation Using Identities
Expanding a 4×4 or 5×5 determinant term by term is painful and error-prone. The smarter route is to transform the determinant into an easy form using properties (the "identities") that change its value in a known, controlled way — then read the answer off a triangular matrix.
The geometric intuition
A determinant measures the signed "volume" of the box spanned by the rows in n-dimensional space. Sliding one row parallel to another doesn't change that volume; swapping two rows flips its sign; scaling a row scales the volume. The algebraic identities are just these facts translated into rules.
The three row (or column) operations
- Swap two rows: det→−det (sign flips).
- Scale a row by k: det→kdet (the factor comes out).
- Add a multiple of one row to a different row (Ri→Ri+λRj, i=j): det unchanged.
The identical rules hold for columns. There is also row-wise linearity: if a row is a sum Ri=Ri′+Ri′′, the determinant splits into the sum of two determinants with all other rows fixed.
Row-wise linearity is not det(A+B)=detA+detB — that is false. The splitting works one row at a time.
The strategy
- Use operation 3 to create zeros in a row or column (value unchanged).
- Factor out common factors with operation 2.
- Swap rows if needed to reach upper-triangular form (track the sign change).
- The determinant is then the product of the diagonal entries.
Worked example
det1472583610.
Apply R2→R2−4R1 and R3→R3−7R1 (no change), then R3→R3−2R2: …
Key idea: expand the determinant, then use A+B+C=0 (so cosC=cos(A+B)) to collapse the trig.
Step 1 — expanding along the first row:
Δ=1−cos2A−cos2B−cos2C+2cosAcosBcosC.
Step 2 — since A+B+C=0, C=−(A+B) so cosC=cos(A+B). Then …
Expanding gives Δ=1−cos2A−cos2B−cos2C+2cosAcosBcosC, and the condition A+B+C=0 collapses the C-terms to cos2A−sin2B, leaving 1−cos2B−sin2B=0.
Intuition
The determinant is a symmetric expression in cosA,cosB,cosC. Expanding it produces a well-known combination, and the single condition A+B+C=0 is exactly what is needed to make that combination vanish, because it lets us write cosC=cos(A+B).
Setting up
Δ=1cosCcosBcosC1cosAcosBcosA1.
Working the steps
1. Expand along the first row:
Δ=1(1−cos2A)−cosC(cosC−cosAcosB)+cosB(cosAcosC−cosB).
Collecting terms,
Δ=1−cos2A−cos2B−cos2C+2cosAcosBcosC.
2. Use the condition. From A+B+C=0 we get C=−(A+B), so cosC=cos(A+B)=cosAcosB−sinAsinB. Group the terms containing cosC:
−cos2C+2cosAcosBcosC=cosC(2cosAcosB−cosC). …
Method: Direct Expansion + Trigonometric Substitution for Conditional Identities
Some determinant identities are proved not by row/column tricks but by expanding the determinant fully first, then using a given condition on the angles (like A+B+C=0 or A+B+C=π) to collapse the resulting trigonometric expression to a known value.
Steps
Step 1: Expand the determinant completely
For a symmetric 3×3 determinant whose entries are cosines (or a mix of 1's and cosines), expand along any row using the standard cofactor formula. The result is a fixed algebraic combination of the cosine terms — usually something of the form 1−cos2A−cos2B−cos2C+2cosAcosBcosC.
Step 2: Bring in the given angle condition
Use the stated relationship between the angles (e.g. C=−(A+B) when A+B+C=0) to rewrite one of the cosines, typically cosC, in terms of the other two using the compound-angle formula cos(A+B)=cosAcosB−sinAsinB.
Step 3: Group and simplify using product-to-sum or Pythagorean identities …
Common Mistakes
Mistake 1: Dropping the minus sign in cos(A+B)
Using A+B+C=0 gives cosC=cos(A+B)=cosAcosB−sinAsinB. Writing this with a + instead of a − breaks the later simplification to cos(A+B)cos(A−B) and the whole proof stalls.
Mistake 2: Sign error in the cofactor expansion
Expanding the 3×3 determinant along the first row, the middle term carries a −cosC cofactor. Missing that sign changes the expanded form 1−cos2A−cos2B−cos2C+2cosAcosBcosC into something that will never reduce to 0. …
Showing the 12 most recent of 14 on this concept.
- GUJCET 2020Set 071 markMCQQ.For △ABC, the value of 0−sin(B+C)tan(A+C)sinA0−cosCtanBcosC0= ________. (A) −1 (B) 0 (C) 1 (D) sinAcosC
›Reveal solutionSolution
Using A+B+C=π the matrix becomes skew-symmetric; a 3×3 (odd-order) skew-symmetric determinant is always 0.
Concept — trig identities in a triangle. Since A+B+C=π: sin(B+C)=sin(π−A)=sinA and tan(A+C)=tan(π−B)=−tanB.
Substituting, the matrix is …
- GUJCET 2025Set 031 markMCQQ.cos2θsin2θ−sin2θcos2θ= _____. (A) 21−21cos22θ (B) 41(3+cos4θ) (C) 1+21sin22θ (D) 1+2sin2θ⋅cos2θ
›Reveal solutionSolution
Expand the 2×2 determinant, then use double/quadruple-angle identities.
cos2θsin2θ−sin2θcos2θ=cos4θ+sin4θ=1−2sin2θcos2θ=1−21sin22θ.
Using sin22θ=21−cos4θ: …
- GUJCET 2022Set 081 markMCQQ.For real numbers x,y,z such that x=y=z, xyzx2y2z21+x31+y31+z3=0 and 111xyzx2y2z2=0 then xyz= ______. (A) 2 (B) −1 (C) 0 (D) 1
›Reveal solutionSolution
Split the last column into 1 and x³; the determinant factors as (Vandermonde)(1+xyz).
Concept. xyzx2y2z21+x31+y31+z3=xyzx2y2z2111+xyzx2y2z2x3y3z3. …
- GUJCET 2020Set 071 markMCQQ.If x,y∈R and (ax+a−x)2(bx+b−x)2(cx+c−x)2(ax−a−x)2(bx−b−x)2(cx−c−x)2111=2y+6 then y= ________. (A) 0 (B) 3 (C) −3 (D) 6
›Reveal solutionSolution
Column 1 − Column 2 =4×(Column 3), so the determinant is 0; 2y+6=0 gives y=−3.
Concept — spot the linear dependence. For any base a, let p=ax, q=a−x, so pq=axa−x=1. Then
(ax+a−x)2−(ax−a−x)2=4axa−x=4
This holds for every row (with bases a,b,c). So in the matrix,
C1−C2=4=4C3 …
- GUJCET 2023Set 091 markMCQQ.sin3611πsin92πcos3611πcos92π= ______. (A) cos12π (B) sin92π (C) cos125π (D) sin127π
›Reveal solutionSolution
A determinant of this sin/cos form collapses to a single sine of the angle difference.
Concept. sinPsinQcosPcosQ=sinPcosQ−cosPsinQ=sin(P−Q).
Solution. P=3611π, Q=92π=368π.
sin(3611π−368π)=sin363π=sin12π. …
- GUJCET 2019Set 171 markMCQQ.sin2θ−cos2θcos2θsin2θ=. (A) 21(1+cos22θ) (B) 21(1−sin22θ) (C) cos2θ (D) 21sin22θ
›Reveal solutionSolution
sin2θ−cos2θcos2θsin2θ=sin4θ+cos4θ, which equals 21(1+cos22θ).
Concept: Expand: sin2θ⋅sin2θ−cos2θ⋅(−cos2θ)=sin4θ+cos4θ.
Now sin4θ+cos4θ=1−2sin2θcos2θ=1−21sin22θ. Using sin22θ=1−cos22θ: …
- GUJCET 2020Set 071 markMCQQ.Let f(t)=cost2tanttantttt12tt. Then limt→0t2f(t) is equal to ________. (A) 3 (B) 1 (C) −1 (D) 0
›Reveal solutionSolution
The determinant equals t(−tcost+tant), so f(t)/t2=−cost+ttant→−1+1=0.
Concept — simplify the determinant first. Column 2 is t[1,1,1]T, so pull out t:
f(t)=tcost2tanttant11112tt=tg(t)
Expand g(t) along the first column's cofactors (about row 1):
g(t)=cost(t−2t)−1(2ttant−2ttant)+1(2tant−tant)
=−tcost+0+tant
Therefore …
- GUJCET 2024Set 131 markMCQQ.If 2017201920182020+2021202320222024=2k, then k3= __________. (A) −64 (B) −8 (C) 0 (D) 8
›Reveal solutionSolution
Both determinants evaluate to −2; their sum −4=2k gives k=−2 and k3=−8.
Concept. Evaluate each 2×2 determinant ad−bc.
Steps.
2017201920182020=2017⋅2020−2018⋅2019=−2, …
- GUJCET 2021Set 151 markMCQQ.For 21−130−2574, the sum of minor and cofactor of 7=. (A) 0 (B) 2 (C) −2 (D) −1
›Reveal solutionSolution
Element 7 sits at position (2,3); cofactor =(−1)2+3× minor.
Concept: Deleting row 2 and column 3:
M23=2−13−2=2(−2)−3(−1)=−1. …
- GUJCET 2019Set 171 markMCQQ.If 1!2!3!2!3!4!3!4!5!=2016K, then K=. (A) 84 (B) 241 (C) 24 (D) 841
›Reveal solutionSolution
Evaluating the determinant of factorials gives 24; with 24=2016K, K=841.
Concept: Write out the values: 1!=1,2!=2,3!=6,4!=24,5!=120. …
- GUJCET 2019Set 171 markMCQQ.Matrix Ar=[rr−1r−1r]; r=1,2,3,… If ∑r=1100∣Ar∣=(10)K, then K=; (∣Ar∣=det(Ar)). (A) 6 (B) 4 (C) 2 (D) 8
›Reveal solutionSolution
∣Ar∣=r2−(r−1)2=2r−1; the sum of the first 100 odd numbers is 1002=10000=(10)8, so K=8.
Concept: ∣Ar∣=rr−1−˚1r=r2−(r−1)2=2r−1. …
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.If f(θ)=cosθsinθ−sinθ−cosθ, then f(6π)= ______.(a) −21(b) 21(c) 23(d) −23
›Reveal solutionSolution
Evaluate the 2×2 determinant, simplify with a double-angle identity, then substitute.
f(θ)=cosθ(−cosθ)−(−sinθ)(sinθ)=−cos2θ+sin2θ=−cos2θ.
…
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.