Properties of Triangle Exercise: 7.1 Class 11 Basic Mathematics Solution [NEB UPDATED]

Exercise 7.1

In any triangle, prove that (Exs. 1-5)

1.a) a(b.cosc – c.cosB) = b² – c²= b² – c² 

Solution:

L.H.S. = a(b.cosc – c.cosB) = b² – c²

= a $\left( {{\rm{b}}.\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}} - {\rm{c}}.\frac{{{{\rm{c}}^2} + {{\rm{a}}^2} - {{\rm{b}}^2}}}{{2{\rm{ca}}}}} \right)$

= $\frac{{\rm{a}}}{{2{\rm{a}}}}$ (a² + b² – c² – c² – a² + b²)

= $\frac{1}{2}$ (2b² – 2c²) = b² – c² = R.H.S. 

b) $\frac{{{\rm{cosA}}}}{{\rm{a}}}$ + $\frac{{\rm{a}}}{{{\rm{bc}}}}$ =$\frac{{{\rm{cosB}}}}{{\rm{b}}}$ + $\frac{{\rm{b}}}{{{\rm{ca}}}}$ = $\frac{{{\rm{cosC}}}}{{\rm{c}}}$ + $\frac{{\rm{c}}}{{{\rm{ab}}}}$

Solution:

L.H.S. = $\frac{{{\rm{cosA}}}}{{\rm{a}}}$ + $\frac{{\rm{a}}}{{{\rm{bc}}}}$ = $\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{{\rm{a}}.2{\rm{bc\:}}}}$ + $\frac{{\rm{a}}}{{{\rm{bc}}}}$

= $\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2} + 2{{\rm{a}}^2}}}{{2{\rm{abc}}}}$ = $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}}}{{2{\rm{abc}}}}$

Middle term =$\frac{{{\rm{cosB}}}}{{\rm{b}}}$ + $\frac{{\rm{b}}}{{{\rm{ca}}}}$ = $\frac{{{{\rm{c}}^2} + {{\rm{a}}^2} - {{\rm{b}}^2}}}{{{\rm{b}}.2{\rm{ac\:}}}}$ + $\frac{{\rm{b}}}{{{\rm{ca}}}}$.

= $\frac{{{{\rm{c}}^2} + {{\rm{a}}^2} - {{\rm{b}}^2} + 2{{\rm{b}}^2}}}{{2{\rm{abc}}}}$ = $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}}}{{2{\rm{abc}}}}$

R.H.S. = $\frac{{{\rm{cosC}}}}{{\rm{c}}}$ + $\frac{{\rm{c}}}{{{\rm{ab}}}}$ = $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{{\rm{c}}.2{\rm{ab\:}}}}$ + $\frac{{\rm{c}}}{{{\rm{ab}}}}$.

= $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2} - 2{{\rm{c}}^2}}}{{2{\rm{abc}}}}$ = $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}}}{{2{\rm{abc}}}}$ 

c) $\frac{{{\rm{cosA}}}}{{\rm{a}}} + \frac{{{\rm{cosB}}}}{{\rm{b}}} + \frac{{{\rm{cosC}}}}{{\rm{c}}}$= $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}}}{{2{\rm{abc}}}}$

Solution:

L.H.S. = $\frac{{{\rm{cosA}}}}{{\rm{a}}} + \frac{{{\rm{cosB}}}}{{\rm{b}}} + \frac{{{\rm{cosC}}}}{{\rm{c}}}$

= $\frac{1}{{\rm{q}}}$. $\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{2{\rm{bc}}}}$ + $\frac{1}{{\rm{b}}}$$\frac{{{{\rm{c}}^2} + {{\rm{a}}^2} - {{\rm{b}}^2}}}{{2{\rm{ca}}}}$ + $\frac{1}{{\rm{c}}}{\rm{\:}}$$\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}}$

= $\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2} + {{\rm{c}}^2} + {{\rm{a}}^2} - {{\rm{b}}^2} + {{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{abc}}}}$

= $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}}}{{2{\rm{abc}}}}$ = R.H.S. 

 

2.a) a³(sin³B – sin³C) + b³(sin³C – sin³A) + c³(sin³A – sin³B)=0

Solution: a³(sin³B – sin³C) + b³(sin³C – sin³A) + c³(sin³A – sin³B)

= a³$\left( {\frac{{{{\rm{b}}^3}}}{{8{{\rm{R}}^3}}} - \frac{{{{\rm{c}}^3}}}{{8{{\rm{r}}^3}}}} \right)$ + b³$\left( {\frac{{{{\rm{c}}^3}}}{{8{{\rm{R}}^3}}} - \frac{{{{\rm{a}}^3}}}{{8{{\rm{r}}^3}}}} \right)$ + c³$\left( {\frac{{{{\rm{a}}^3}}}{{8{{\rm{R}}^3}}} - \frac{{{{\rm{b}}^3}}}{{8{{\rm{r}}^3}}}} \right)$

= $\frac{1}{{8{{\rm{r}}^3}}}$(a³b³ – a³c³ + b³c³ – a³b³ + c³a³ – b³c³)z = $\frac{1}{{8{{\rm{r}}^3}}}$. 0

= 0 = R.H.S. 

b) $\frac{{{{\rm{a}}^2}\sin \left( {{\rm{B}} - {\rm{c}}} \right)}}{{{\rm{sinA}}}}$ + $\frac{{{{\rm{b}}^2}{\rm{sin}}\left( {{\rm{C}} - {\rm{A}}} \right)}}{{{\rm{sinB}}}}$ + $\frac{{{{\rm{c}}^2}\sin \left( {{\rm{A}} - {\rm{B}}} \right)}}{{{\rm{sinC}}}}$ =0

Solution:

L.H.S. = $\frac{{{{\rm{a}}^2}\sin \left( {{\rm{B}} - {\rm{c}}} \right)}}{{{\rm{sinA}}}}$ + $\frac{{{{\rm{b}}^2}{\rm{sin}}\left( {{\rm{C}} - {\rm{A}}} \right)}}{{{\rm{sinB}}}}$ + $\frac{{{{\rm{c}}^2}\sin \left( {{\rm{A}} - {\rm{B}}} \right)}}{{{\rm{sinC}}}}$

= $\frac{{4{{\rm{R}}^2}{{\sin }^2}{\rm{A}}.\sin \left( {{\rm{B}} - {\rm{C}}} \right)}}{{{\rm{sinA}}}}$ + $\frac{{4{{\rm{R}}^2}{{\sin }^2}{\rm{B}}.\sin \left( {{\rm{C}} - {\rm{A}}} \right)}}{{{\rm{sinB}}}}$ + $\frac{{4{{\rm{R}}^2}{{\sin }^2}{\rm{C}}.\sin \left( {{\rm{A}} - {\rm{B}}} \right)}}{{{\rm{sinC}}}}$

= 4R² {sinA.sin(B – C) + sinB.sin(C – A) + sinC sin(A – B)}

= 4R² {sin(B+C).sin(B – C) + sin(C + A).sin(C – A) + sin(A + B).sin(A – B)}   [A + B + C = π]

= 4R² (sin²B – sin²C + sin²C – sin²A + sin²A – sin²B)             [sin(x+y).sin(x – y) = sin²x. – sin²y]

= 4R². 0 = 0 = R.H.S.

 c) $\frac{{{\rm{asin}}\left( {{\rm{B}} - {\rm{C}}} \right)}}{{{{\rm{b}}^2} - {{\rm{c}}^2}}}$=$\frac{{{\rm{bsin}}\left( {{\rm{C}} - {\rm{A}}} \right)}}{{{{\rm{c}}^2} - {{\rm{a}}^2}}}$= $\frac{{{\rm{C}}.\sin \left( {{\rm{A}} - {\rm{B}}} \right)}}{{{{\rm{a}}^2} - {{\rm{b}}^2}}}$

Solution:

L.H.S. = $\frac{{{\rm{asin}}\left( {{\rm{B}} - {\rm{C}}} \right)}}{{{{\rm{b}}^2} - {{\rm{c}}^2}}}$ = $2{\rm{R}}.{\rm{sinA}}.\frac{{\sin \left( {{\rm{B}} - {\rm{C}}} \right)}}{{{{\rm{b}}^2} - {{\rm{c}}^2}}}$

= $\frac{{2{\rm{R}}.\sin \left( {{\rm{B}} + {\rm{C}}} \right).\sin \left( {{\rm{B}} - {\rm{C}}} \right)}}{{{{\rm{b}}^2} - {{\rm{c}}^2}}}$   [B+C = π – A]

= $\frac{{2{\rm{R}}.\left( {{{\sin }^2}{\rm{B}} - {{\sin }^2}{\rm{C}}} \right)}}{{4{{\rm{R}}^2}\left( {{{\sin }^2}{\rm{B}} - {{\sin }^2}{\rm{C}}} \right)}}$   [b = 2R sinB etc.]

= $\frac{1}{{2{\rm{R}}}}$.

Middle term = $\frac{{{\rm{bsin}}\left( {{\rm{C}} - {\rm{A}}} \right)}}{{{{\rm{c}}^2} - {{\rm{a}}^2}}}$ = $\frac{{2{\rm{RsinB}}.\sin \left( {{\rm{C}} - {\rm{A}}} \right)}}{{{{\rm{c}}^2} - {{\rm{a}}^2}}}$

= $\frac{{2{\rm{R}}.\sin \left( {{\rm{C}} + {\rm{A}}} \right).\sin \left( {{\rm{C}} - {\rm{A}}} \right)}}{{{{\rm{c}}^2} - {{\rm{a}}^2}}}$  [C + A = π – B]

= $\frac{{2{\rm{R}}.\left( {{{\sin }^2}{\rm{C}} - {{\sin }^2}{\rm{A}}} \right)}}{{4{{\rm{R}}^2}\left( {{{\sin }^2}{\rm{C}} - {{\sin }^2}{\rm{A}}} \right)}}$  [C = 2R sinC .etc]

= $\frac{1}{{2{\rm{R}}}}$

R.H.S. = $\frac{{{\rm{C}}.\sin \left( {{\rm{A}} - {\rm{B}}} \right)}}{{{{\rm{a}}^2} - {{\rm{b}}^2}}}$ = $\frac{{2{\rm{RsinC}}.\sin \left( {{\rm{A}} - {\rm{B}}} \right)}}{{{{\rm{a}}^2} - {{\rm{b}}^2}}}$

= $\frac{{2{\rm{R}}.\sin \left( {{\rm{A}} + {\rm{B}}} \right).\sin \left( {{\rm{A}} - {\rm{B}}} \right)}}{{{{\rm{a}}^2} - {{\rm{b}}^2}}}$  [A + B = π – C]

= $\frac{{2{\rm{R}}.\left( {{{\sin }^2}{\rm{A}} - {{\sin }^2}{\rm{B}}} \right)}}{{4{{\rm{R}}^2}\left( {{{\sin }^2}{\rm{A}} - {{\sin }^2}{\rm{B}}} \right)}}$   [a = 2R sinA etc.]

= $\frac{1}{{2{\rm{R}}}}$

Hence, L.H.S. = Middle term = R.H.S.

 

3.a) (b + c)cosA + (c + a)cosB + (a + B)cosC=a+b+c

Solution:

L.H.S. = (b + c)cosA + (c + a)cosB + (a + B)cosC

= b.cosA + c.cosA + c.cosB + a.cosB + a.cosC + b.cosC

= (bcosA + acosB) + (ccosA +a.cosC) + (b.cosC + c.cosB)

= c + b + a   [ a= b.cosC + c.cosB]

= a + b + c = R.H.S. 

b) b²sin2C + c² sin2B=2ab.sinC

Solution:

L.H.S. = b²sin2C + c² sin2B

= b². 2sinC.cosC + c².2sinB.cosB

= 2b² . $\frac{{\rm{c}}}{{2{\rm{R}}}}$ . $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}}$ + 2c² . $\frac{{\rm{b}}}{{2{\rm{R}}}}$ .$\frac{{{{\rm{c}}^{\rm{a}}} + {{\rm{a}}^2} - {{\rm{b}}^2}}}{{2{\rm{ca}}}}$

= $\frac{{{\rm{bc}}}}{{2{\rm{Ra}}}}$. $\left( {{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}} \right)$ + $\frac{{{\rm{bc}}}}{{2{\rm{Ra}}}}$. $\left( {{{\rm{c}}^2} + {{\rm{a}}^2} - {{\rm{b}}^2}} \right)$

= $\frac{{\rm{b}}}{{2{\rm{Ra}}}}\left( {{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2} + {{\rm{c}}^2} + {{\rm{a}}^2} - {{\rm{b}}^2}} \right)$

= $\frac{{{\rm{bc}}}}{{2{\rm{Ra}}}}$.2a² = $\frac{{2{\rm{abc}}}}{{2{\rm{R}}}}$ = 2ab. $\frac{{\rm{c}}}{{2{\rm{R}}}}$ = 2ab.sinC = R.H.S.

c) c²cos²B + b²cos²B + bc.cos(B – C) = $\frac{1}{2}\left( {{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}} \right)$

Solution:

L.H.S. = c²cos²B + b²cos²B + bc.cos(B – C)

= (c.cosB + b.cosC)² – 2bc.cosB.cosC + bc.cosB.cosC +bc.sinB.sinC

= a² – bc(cosB.cosC – sinB.sinC)     [a = b.cosC + cosB]

= a² – bc.cos(B + C)

= a² – bc.$\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{2{\rm{bc}}}}$

=$\frac{{2{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{2}$

= $\frac{1}{2}\left( {{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}} \right)$ = R.H.S.

d) $\frac{{{\rm{c}} - {\rm{bcosA}}}}{{{\rm{b}} - {\rm{c}}.{\rm{cosA}}}}$ =$\frac{{{\rm{CosB}}}}{{{\rm{cosC}}}}$

Solution:

L.H.S. = $\frac{{{\rm{c}} - {\rm{bcosA}}}}{{{\rm{b}} - {\rm{c}}.{\rm{cosA}}}}$

= $\frac{{{\rm{a}}.{\rm{cosB}} + {\rm{b}}.{\rm{cosA}} - {\rm{bcosA}}}}{{{\rm{acosC}} + {\rm{c}}.{\rm{cosA}} - {\rm{c}}.{\rm{cosA}}}}$   [b = acosC + c.cosA]

= $\frac{{{\rm{a}}.{\rm{cosB}}}}{{{\rm{a}}.{\rm{cosC}}}}$ = $\frac{{{\rm{CosB}}}}{{{\rm{cosC}}}}$ = R.H.S.

e) $\frac{{{\rm{a}} - {\rm{bcosC}}}}{{{\rm{c}} - {\rm{b}}.{\rm{cosA}}}}$=$\frac{{{\rm{sinC}}}}{{{\rm{sinA}}}}$

Solution:

L.H.S. = $\frac{{{\rm{a}} - {\rm{bcosC}}}}{{{\rm{c}} - {\rm{b}}.{\rm{cosA}}}}$

= $\frac{{{\rm{b}}.{\rm{cosC}} + {\rm{c}}.{\rm{cosC}} - {\rm{bcosC}}}}{{{\rm{acosB}} + {\rm{b}}.{\rm{cosA}} - {\rm{b}}.{\rm{cosA}}}}$   [a = bcosC + c.cosB.etc]

= $\frac{{{\rm{c}}.{\rm{cosB}}}}{{{\rm{a}}.{\rm{cosB}}}}$ = $\frac{{2{\rm{R}}.{\rm{sinC}}}}{{2{\rm{RsinA}}}}$ = $\frac{{{\rm{sinC}}}}{{{\rm{sinA}}}}$ = R.H.S.

 

4.a) a³.sin(B – C) + b³.sin(C – A) + c³.sin(A – B) =0

L.H.S. = a³.sin(B – C) + b³.sin(C – A) + c³.sin(A – B)

= a².a.sin(B – C) + b².b sin(C – A) + c².c .sin(A – B)

= a².2R sinA.sin(B – C) + b².2R sinB.sin(C – A) + c².2R.sinC.sin(A – B)          [a = 2R sinA.etc]

= 2Ra² sin(B + C)sin(B – C) + 2Rb² sin(C + A).sin(C – A) + 2Rc² sin(A + B).sin(A – B).

= 2R{a²(sin²B – sin²C) + b²(sin²C – sin²A) + c² (sin²A – sin²B)}

= 2R $\{ {{\rm{a}}^2}\left( {\frac{{{{\rm{b}}^2}}}{{4{{\rm{R}}^2}}} - \frac{{{{\rm{c}}^2}}}{{4{{\rm{R}}^2}}}} \right) + {{\rm{b}}^2}\left( {\frac{{{{\rm{c}}^2}}}{{4{{\rm{R}}^2}}} - \frac{{{{\rm{a}}^2}}}{{4{{\rm{R}}^2}}}} \right) + {{\rm{c}}^2}\left( {\frac{{{{\rm{a}}^2}}}{{4{{\rm{R}}^2}}} - \frac{{{{\rm{b}}^2}}}{{4{{\rm{R}}^2}}}} \right)$

= $\frac{{2{\rm{R}}}}{{4{{\rm{R}}^2}}}$ (a²b² – a²c² + b²c² – a²b² + c²a² – b²c²) = $\frac{1}{{2{\rm{R}}}}$. 0 = 0 = R.H.S.

 

b) $\frac{{{{\rm{b}}^2} - {{\rm{c}}^2}}}{{{{\rm{a}}^2}}}$ sin2A + $\frac{{{{\rm{c}}^2} - {{\rm{a}}^2}}}{{{{\rm{b}}^2}}}$sin2B + $\frac{{{{\rm{a}}^2} - {{\rm{b}}^2}}}{{{{\rm{c}}^2}}}$ sin2C=0

Solution:

L.H.S. = $\frac{{{{\rm{b}}^2} - {{\rm{c}}^2}}}{{{{\rm{a}}^2}}}$ sin2A + $\frac{{{{\rm{c}}^2} - {{\rm{a}}^2}}}{{{{\rm{b}}^2}}}$sin2B + $\frac{{{{\rm{a}}^2} - {{\rm{b}}^2}}}{{{{\rm{c}}^2}}}$ sin2C

= $\frac{{{{\rm{b}}^2} - {{\rm{c}}^2}}}{{{{\rm{a}}^2}}}$. 2sinA.cosA + $\frac{{{{\rm{c}}^2} - {{\rm{a}}^2}}}{{{{\rm{b}}^2}}}$. 2sinB.cosB + $\frac{{{{\rm{a}}^2} - {{\rm{b}}^2}}}{{{{\rm{c}}^2}}}$.2sinC.cosC

= $\frac{{2\left( {{{\rm{b}}^2} - {{\rm{c}}^2}} \right)}}{{{{\rm{a}}^2}}}$. $\frac{{\rm{a}}}{{2{\rm{R}}}}$ . $\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{2{\rm{bc}}}}$ +  $\frac{{2\left( {{{\rm{c}}^2} - {{\rm{a}}^2}} \right)}}{{{{\rm{b}}^2}}}$. $\frac{{\rm{b}}}{{2{\rm{R}}}}$ . $\frac{{{{\rm{c}}^2} + {{\rm{a}}^2} - {{\rm{b}}^2}}}{{2{\rm{ca}}}}$ + $\frac{{2\left( {{{\rm{a}}^2} - {{\rm{b}}^2}} \right)}}{{{{\rm{c}}^2}}}$. $\frac{{\rm{c}}}{{2{\rm{R}}}}$ . $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}}$

 [$\frac{{\rm{a}}}{{{\rm{sinA}}}}$ = 2R. etc. and cosA = $\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{2{\rm{bc}}}}$]

= $\frac{1}{{2{\rm{abcR}}}}$ {(b² – c²)(b² + c² – a²) + (c² – a²)(c²+a² – b²) + (a² – b²)(a² + b² – c²)}

= $\frac{1}{{2{\rm{abcR}}}}$[b⁴ – c⁴ – a²(b² – c²) + c⁴ – a⁴ – b²(c² – a²) + a⁴ – b⁴ – c²(a² – b²)}

= $\frac{1}{{2{\rm{abcR}}}}$ [–a²b² + a²c² – b²c² + a²b² – c²a² + b²c²}

= $\frac{1}{{2{\rm{abcR}}}}$ . 0 = 0 = R.H.S. 

c) sin(A + B) : sin(A – B) = c²:(a²-b²)

Solution:

L.H.S. = sin(A + B) : sin(A – B)

= $\frac{{\sin \left( {{\rm{A}} + {\rm{B}}} \right)}}{{\sin \left( {{\rm{A}} - {\rm{B}}} \right)}}$ = $\frac{{\sin \left( {{\rm{A}} + {\rm{B}}} \right)}}{{\sin \left( {{\rm{A}} - {\rm{B}}} \right)}}.\frac{{\sin \left( {{\rm{A}} + {\rm{B}}} \right)}}{{\sin \left( {{\rm{A}} + {\rm{B}}} \right)}}$

= $\frac{{{\rm{sinC}}.{\rm{sinC}}}}{{{{\sin }^2}{\rm{A}} - {{\sin }^2}{\rm{B}}}}$     [ A + B = π – C and sin(A+B) .sin(A – B) = sin²A – sin²B]

= $\frac{{{{\sin }^2}{\rm{C}}}}{{{{\sin }^2}{\rm{A}} - {{\sin }^2}{\rm{B}}}}$ = $\frac{{{{\rm{c}}^2}}}{{4{{\rm{R}}^2}\left( {\frac{{{{\rm{a}}^2}}}{{4{{\rm{R}}^2}}} - \frac{{{{\rm{b}}^2}}}{{4{{\rm{R}}^2}}}} \right)}}$  [sinA = $\frac{{\rm{a}}}{{2{\rm{R}}}}$,etc.]

= $\frac{{{{\rm{c}}^2}}}{{{{\rm{a}}^2} - {{\rm{b}}^2}}}$ = R.H.S. 

d) (a² – b² + c²) tanB =(a² + b² – c²) tanC=(b² + c² – a²) tanA

Solution:

L.H.S. = (a² – b² + c²) tanB = (a² + c² – b²) $\frac{{{\rm{sinB}}}}{{{\rm{cosB}}}}$.

= (a² + c² – b²). $\frac{{\rm{b}}}{{2{\rm{R}}}}$$\left( {\frac{{2{\rm{ac}}}}{{{{\rm{a}}^2} + {{\rm{c}}^2} - {{\rm{b}}^2}}}} \right)$ = $\frac{{{\rm{abc}}}}{{\rm{R}}}{\rm{\:}}$ 

Middle term = (a² + b² – c²) tanC = (a² + b² – c²) $\frac{{{\rm{sinC}}}}{{{\rm{cosC}}}}$.

= (a² + b² – c²) $\frac{{\rm{c}}}{{2{\rm{R}}}}$$\left( {\frac{{2{\rm{ab}}}}{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}} \right)$ = $\frac{{{\rm{abc}}}}{{\rm{R}}}$ 

R.H.S. = (b² + c² – a²) tanA = (b² + c² – a²) $\frac{{{\rm{SinA}}}}{{{\rm{cosA}}}}$.

= (b² + c² – a²) $\frac{{\rm{a}}}{{2{\rm{R}}}}$$\left( {\frac{{2{\rm{ab}}}}{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}} \right)$ = $\frac{{{\rm{abc}}}}{{\rm{R}}}$.

Hence, L.H.S. = Middle term = R.H.S. 

e) $\frac{{{\rm{cosB}} - {\rm{cosC}}}}{{{\rm{cosA}} + 1}}$ = $\frac{{{\rm{c}} - {\rm{b}}}}{{\rm{a}}}$

Solution:

L.H.S. = $\frac{{{\rm{cosB}} - {\rm{cosC}}}}{{{\rm{cosA}} + 1}}$ = $\frac{{\left( {2{{\cos }^2}\frac{{\rm{B}}}{2} - 1} \right) - \left( {2{{\cos }^2}\frac{{\rm{C}}}{2} - 1} \right)}}{{2{{\cos }^2}\frac{{\rm{A}}}{2}}}$

= $\frac{{{{\cos }^2}\frac{{\rm{B}}}{2} - {{\cos }^2}\frac{{\rm{C}}}{2}}}{{{{\cos }^2}\frac{{\rm{A}}}{2}}}$ = $\frac{{\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{b}}} \right)}}{{{\rm{ac}}}} - \frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{c}}} \right)}}{{{\rm{ab}}}}}}{{\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{a}}} \right)}}{{{\rm{bc}}}}}}$ = $\frac{{{\rm{bs}}\left( {{\rm{s}} - {\rm{b}}} \right) - {\rm{cs}}\left( {{\rm{s}} - {\rm{c}}} \right)}}{{{\rm{as}}\left( {{\rm{s}} - {\rm{a}}} \right)}}$

= $\frac{{{\rm{s}}\left( {{\rm{bs}} - {{\rm{b}}^2} - {\rm{cs}} + {{\rm{c}}^2}} \right)}}{{{\rm{sa}}\left( {{\rm{s}} - {\rm{a}}} \right)}}$ = $\frac{{{\rm{s}}\left( {{\rm{b}} - {\rm{c}}} \right) - ({{\rm{b}}^2} - {{\rm{c}}^2})}}{{{\rm{a}}\left( {{\rm{s}} - {\rm{a}}} \right)}}{\rm{\:\:}}$

= $\frac{{{\rm{s}}\left( {{\rm{b}} - {\rm{c}}} \right) - \left( {{\rm{b}} - {\rm{c}}} \right)\left( {{\rm{b}} + {\rm{c}}} \right)}}{{{\rm{a}}\left( {{\rm{s}} - {\rm{a}}} \right)}}$ = $\frac{{\left( {{\rm{b}} - {\rm{c}}} \right)\left\{ {{\rm{s}} - \left( {{\rm{b}} + {\rm{c}}} \right)} \right\}}}{{{\rm{a}}\left( {{\rm{s}} - {\rm{a}}} \right)}}$

= $\frac{{{\rm{b}} - {\rm{c}}}}{{\rm{a}}}$. $\frac{{\left\{ {{\rm{s}} - \left( {2{\rm{s}} - {\rm{a}}} \right)} \right\}}}{{{\rm{s}} - {\rm{a}}}}$ = $\frac{{{\rm{b}} - {\rm{c}}}}{{\rm{a}}}$. $\frac{{ - {\rm{s}} + {\rm{a}}}}{{{\rm{s}} - {\rm{a}}}}$ = $\frac{{{\rm{c}} - {\rm{b}}}}{{\rm{a}}}$ = R.H.S.

f) $\frac{{{\rm{asinA}} + {\rm{bsinB}} + {\rm{csinC}}}}{{{\rm{acosA}} + {\rm{bcosB}} + {\rm{c}}.{\rm{cosC}}}}$ = $\frac{{\rm{R}}}{{{\rm{abc}}}}$(a² + b² + c²)

Solution:

L.H.S. = $\frac{{{\rm{asinA}} + {\rm{bsinB}} + {\rm{csinC}}}}{{{\rm{acosA}} + {\rm{bcosB}} + {\rm{c}}.{\rm{cosC}}}}$

= $\frac{{{\rm{a}}.\frac{{\rm{a}}}{{2{\rm{R}}}} + {\rm{b}}.\frac{{\rm{b}}}{{2{\rm{R}}}} + {\rm{c}}.\frac{{\rm{c}}}{{2{\rm{R}}}}}}{{2{\rm{RsinA}}.{\rm{cosA}} + 2{\rm{RsinB}}.{\rm{cosB}} + 2{\rm{RsinC}}.{\rm{cosC}}}}$

= $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}}}{{2{\rm{R}}.{\rm{R}}\left( {2{\rm{sinA}}.{\rm{cosA}} + 2{\rm{sinB}}.{\rm{cosB}} + 2{\rm{sinC}}.{\rm{cosC}}} \right)}}$

= $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}}}{{2{{\rm{R}}^2}\left( {{\rm{sin}}2{\rm{A}} + {\rm{sin}}2{\rm{B}} + {\rm{sin}}2{\rm{C}}} \right)}}$

= $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}}}{{2{{\rm{R}}^2}.4{\rm{sinA}}.{\rm{sinB}}.{\rm{sinC}}}}$                                      [A+B+ C = π →sin2A + sin2B +sin2C =4sinA.sinB.sinC]

= $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}}}{{8{{\rm{R}}^2}.\frac{{\rm{a}}}{{2{\rm{R}}}}.\frac{{\rm{b}}}{{2{\rm{R}}}}.\frac{{\rm{c}}}{{2{\rm{R}}}}}}$

= $\frac{{\rm{R}}}{{{\rm{abc}}}}$(a² + b² + c²) = R.H.S.

 

5. a) $\frac{{{\rm{b}} - {\rm{c}}}}{{\rm{a}}}$. Cos $\frac{{\rm{A}}}{2}$ = sin $\frac{{{\rm{B}} - {\rm{C}}}}{2}$

Solution:

L.H.S. = $\frac{{{\rm{b}} - {\rm{c}}}}{{\rm{a}}}$. Cos $\frac{{\rm{A}}}{2}$ = $\frac{{2{\rm{RsinB}} - 2{\rm{RsinC}}}}{{2{\rm{RsinA}}}}$. Cos $\frac{{\rm{A}}}{2}$

= $\frac{{{\rm{sinB}} - {\rm{sinC}}}}{{{\rm{sinA}}}}$. Cos $\frac{{\rm{A}}}{2}$ = $\frac{{2\cos \frac{{{\rm{B}} + {\rm{C}}}}{2}.\sin \frac{{{\rm{B}} - {\rm{C}}}}{2}}}{{2\sin \frac{{\rm{A}}}{2}.\cos \frac{{\rm{A}}}{2}}}$. cos $\frac{{\rm{A}}}{2}$

= $\frac{{2\sin \frac{{\rm{A}}}{2}.\sin \frac{{{\rm{B}} - {\rm{C}}}}{2}}}{{2\sin \frac{{\rm{A}}}{2}}}$ = sin $\frac{{{\rm{B}} - {\rm{C}}}}{2}$ = R.H.S. 

b) bcos²$\frac{{\rm{A}}}{2}$ + a.cos²$\frac{{\rm{B}}}{2}$ = $\frac{{{\rm{a}} + {\rm{b}} + {\rm{c}}}}{2}$

Solution:

L.H.S. = bcos²$\frac{{\rm{A}}}{2}$ + a.cos²$\frac{{\rm{B}}}{2}$ = b. $\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{a}}} \right)}}{{{\rm{bc}}}}$ + a. $\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{b}}} \right)}}{{{\rm{ca}}}}$

= $\frac{{{{\rm{s}}^2} - {\rm{as}}}}{{\rm{c}}}$ + $\frac{{{{\rm{s}}^2} - {\rm{bs}}}}{{\rm{c}}}$ = $\frac{{2{{\rm{s}}^2} - \left( {{\rm{a}} + {\rm{b}}} \right){\rm{s}}}}{{\rm{c}}}$

= $\frac{{2{{\rm{s}}^2} - \left( {2{\rm{s}} - {\rm{c}}} \right){\rm{s}}}}{{\rm{c}}}$ = $\frac{{2{{\rm{s}}^2} - 2{{\rm{s}}^2} + {\rm{cs}}}}{{\rm{c}}}$ = $\frac{{{\rm{cs}}}}{{\rm{c}}}$ = s = $\frac{{{\rm{a}} + {\rm{b}} + {\rm{c}}}}{2}$ = R.H.S. 

c) a(cosB – cosC) =2(c –b) cos²$\frac{{\rm{A}}}{2}$

Solution:

L.H.S. = a(cosB – cosC)

= a$\left( {\frac{{{{\rm{c}}^2} + {{\rm{a}}^2} - {{\rm{b}}^2}}}{{2{\rm{ca}}}} - \frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}}} \right)$ = $\frac{{{{\rm{c}}^2} + {{\rm{a}}^2} - {{\rm{b}}^2}}}{{2{\rm{c}}}}$ – $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{b}}}}$

= $\frac{{{\rm{b}}{{\rm{c}}^2} + {{\rm{a}}^2}{\rm{b}} - {{\rm{b}}^3} - {{\rm{a}}^2}{\rm{c}} - {{\rm{b}}^2}{\rm{c}} + {{\rm{c}}^3}}}{{2{\rm{bc}}}}$

= $\frac{{{{\rm{c}}^3} - {{\rm{b}}^3} - {{\rm{b}}^2}{\rm{c}} + {\rm{b}}{{\rm{c}}^2} - {{\rm{a}}^2}{\rm{c}} + {{\rm{a}}^2}{\rm{b}}}}{{2{\rm{bc}}}}$

= $\frac{{\left( {{\rm{c}} - {\rm{b}}} \right)\left( {{{\rm{c}}^2} + {\rm{cb}} + {{\rm{b}}^2}} \right) - {\rm{bc}}\left( {{\rm{b}} - {\rm{c}}} \right) - {{\rm{a}}^2}\left( {{\rm{c}} - {\rm{b}}} \right)}}{{2{\rm{bc}}}}$

= $\frac{{{\rm{c}} - {\rm{b}}}}{{2{\rm{cb}}}}$ (c² + cb + b² + bc – a²)

= $\left( {\frac{{{\rm{c}} - {\rm{b}}}}{{2{\rm{cb}}}}} \right)$(b² + 2bc + c² – a²) = $\frac{{{\rm{c}} - {\rm{b}}}}{{2{\rm{bc}}}}$ {(b + c)² – a²}

= $\left( {\frac{{{\rm{c}} - {\rm{b}}}}{{2{\rm{bc}}}}} \right)$ (b + c + a)(b + c – a)

= $\frac{{\left( {{\rm{c}} - {\rm{b}}} \right).2{\rm{s}}.\left( {2{\rm{s}} - 2{\rm{a}}} \right)}}{{2{\rm{bc}}}}$    [a + b + c = 2s]

= 2(c – b). $\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{a}}} \right)}}{{{\rm{bc}}}}$ = 2(c –b) cos²$\frac{{\rm{A}}}{2}$ = R.H.S. 

d) $\frac{{{\rm{b}} - {\rm{c}}}}{{\rm{a}}}$ cos²$\frac{{\rm{A}}}{2}$ + $\frac{{{\rm{c}} - {\rm{a}}}}{{\rm{b}}}$. Cos²$\frac{{\rm{B}}}{2}$ + $\frac{{{\rm{a}} - {\rm{b}}}}{{\rm{c}}}$ cos²$\frac{{\rm{C}}}{2}$=0

Solution:

L.H.S. = $\frac{{{\rm{b}} - {\rm{c}}}}{{\rm{a}}}$ cos²$\frac{{\rm{A}}}{2}$ + $\frac{{{\rm{c}} - {\rm{a}}}}{{\rm{b}}}$. Cos²$\frac{{\rm{B}}}{2}$ + $\frac{{{\rm{a}} - {\rm{b}}}}{{\rm{c}}}$ cos²$\frac{{\rm{C}}}{2}$

= $\frac{{{\rm{b}} - {\rm{c}}}}{2}$. $\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{a}}} \right)}}{{{\rm{bc}}}}$ + $\frac{{{\rm{c}} - {\rm{a}}}}{{\rm{b}}}$. $\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{b}}} \right)}}{{{\rm{ca}}}}$ + $\frac{{{\rm{a}} - {\rm{b}}}}{{\rm{c}}}$. $\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{c}}} \right)}}{{{\rm{ab}}}}$

= $\frac{{\rm{s}}}{{{\rm{abc}}}}${(b – c)(s – a) + (c – a)(s – b) + (a – b)(s – c)}

= $\frac{{\rm{s}}}{{{\rm{abc}}}}$ {bs – ab – cs + ca + cs – bc – as + ab + as – ac – bs + bc}

= $\frac{{\rm{s}}}{{{\rm{abc}}}}$.0 = 0 = R.H.S. 

e) bc.cos²$\frac{{\rm{A}}}{2}$ + ca.cos²$\frac{{\rm{B}}}{2}$ + ab.cos²$\frac{{\rm{C}}}{2}$= s² 

Solution:

L.H.S. = bc.cos²$\frac{{\rm{A}}}{2}$ + ca.cos²$\frac{{\rm{B}}}{2}$ + ab.cos²$\frac{{\rm{C}}}{2}$

= bc. $\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{a}}} \right)}}{{{\rm{bc}}}}$ + ca. $\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{b}}} \right)}}{{{\rm{ca}}}}$ + ab.$\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{c}}} \right)}}{{{\rm{ab}}}}$

= s² – as + s² – bs + s² – cs = 3s² – s(a + b + c).

= 3s² – s.2s = 3s² – 2s² = s² = R.H.S. 

f) tan²$\frac{{\rm{A}}}{2}$. tan²$\frac{{\rm{B}}}{2}$.tan²$\frac{{\rm{C}}}{2}$=$\left( {\frac{{{\rm{s}} - {\rm{a}}}}{{\rm{s}}}} \right)$.$\left( {\frac{{{\rm{s}} - {\rm{b}}}}{{\rm{s}}}} \right)$.${\rm{\:}}\left( {\frac{{{\rm{s}} - {\rm{c}}}}{{\rm{s}}}} \right)$

Solution:

L.H.S. = tan²$\frac{{\rm{A}}}{2}$. tan²$\frac{{\rm{B}}}{2}$.tan²$\frac{{\rm{C}}}{2}$

= $\frac{{\left( {{\rm{s}} - {\rm{b}}} \right).\left( {{\rm{s}} - {\rm{c}}} \right)}}{{{\rm{s}}\left( {{\rm{s}} - {\rm{a}}} \right)}}$.$\frac{{\left( {{\rm{s}} - {\rm{c}}} \right)\left( {{\rm{s}} - {\rm{a}}} \right)}}{{{\rm{s}}\left( {{\rm{s}} - {\rm{b}}} \right)}}$.${\rm{\:}}\frac{{\left( {{\rm{s}} - {\rm{a}}} \right)\left( {{\rm{s}} - {\rm{b}}} \right)}}{{{\rm{s}}\left( {{\rm{s}} - {\rm{c}}} \right)}}$

= $\left( {\frac{{{\rm{s}} - {\rm{a}}}}{{\rm{s}}}} \right)$.$\left( {\frac{{{\rm{s}} - {\rm{b}}}}{{\rm{s}}}} \right)$.${\rm{\:}}\left( {\frac{{{\rm{s}} - {\rm{c}}}}{{\rm{s}}}} \right)$ = R.H.S. 

g) (b + c – a)(cot $\frac{{\rm{B}}}{2}$ + cot $\frac{{\rm{C}}}{2}$)=2a. cot $\frac{{\rm{A}}}{2}$

Solution:

L.H.S. = (b + c – a)(cot $\frac{{\rm{B}}}{2}$ + cot $\frac{{\rm{C}}}{2}$)

= (2s – 2a) $\left\{ {\sqrt {\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{b}}} \right)}}{{\left( {{\rm{s}} - {\rm{a}}} \right)\left( {{\rm{s}} - {\rm{c}}} \right)}}}  + \sqrt {\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{c}}} \right)}}{{\left( {{\rm{s}} - {\rm{a}}} \right)\left( {{\rm{s}} - {\rm{b}}} \right)}}} } \right\}$

= 2(s – a) $\frac{{\sqrt {\rm{s}} }}{{\sqrt {{\rm{s}} - {\rm{a}}} }}$$\left( {\frac{{{\rm{s}} - {\rm{b}} + {\rm{s}} - {\rm{c}}}}{{\sqrt {\left( {{\rm{s}} - {\rm{b}}} \right)\left( {{\rm{s}} - {\rm{c}}} \right)} }}} \right)$

= 2$\sqrt {\left( {{\rm{s}} - {\rm{a}}} \right){\rm{s}}} $$\frac{{\rm{a}}}{{\sqrt {\left( {{\rm{s}} - {\rm{b}}} \right).\left( {{\rm{s}} - {\rm{c}}} \right)} }}$    [2s – b – c = a]

= 2a $\sqrt {\frac{{\left( {{\rm{s}}\left( {{\rm{s}} - {\rm{a}}} \right)} \right)}}{{\left( {{\rm{s}} - {\rm{b}}} \right)\left( {{\rm{s}} - {\rm{c}}} \right)}}} $ = 2a. cot $\frac{{\rm{A}}}{2}$ = R.H.S.

 

6. If a⁴ + b⁴ + c⁴ = 2c² (a² + b²), prove that C=45⁰ or 135⁰.

Solution:

Or, a⁴ + b⁴ + c⁴ = 2c² (a² + b²)

Or, (a² + b²)² – 2a²b² – 2c²(a² + b²) + c⁴ = 0

Or, (a² + b²)² – 2.(a² + b²).c² + (c²)² = 2a².b².

Or, (a² + b² – c²)² = 2a²b²

Or, ${\left( {\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}}} \right)^2}$ = $\frac{1}{2}$.

Or, cos²C = $\frac{1}{2}$

Or, cos C = $ \pm \frac{1}{{\sqrt 2 }}$

Hence, C = 45° or 135°.

 

7) If (a + b + c)(b + c – a) = 3bc, show that A=60⁰.

Solution:

Here, (a + b + c)(b + c – a) = 3bc

Or, (b + c)² – a² = 3bc

Or, b² + 2bc + c² – a² = 3bc

Or, b² + c² – a² = bc.

Or, $\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{2{\rm{bc}}}}$ = $\frac{1}{2}$.

Or, cosA = $\frac{1}{2}$

So, A = 60°

 

8) If $\frac{1}{{{\rm{a}} + {\rm{C}}}}$ + $\frac{1}{{{\rm{b}} + {\rm{c}}}}$ = $\frac{3}{{{\rm{a}} + {\rm{b}} + {\rm{c}}}}$, show that C=60⁰.

Solution:

Here, $\frac{1}{{{\rm{a}} + {\rm{C}}}}$ + $\frac{1}{{{\rm{b}} + {\rm{c}}}}$ = $\frac{3}{{{\rm{a}} + {\rm{b}} + {\rm{c}}}}$

Or, $\frac{{{\rm{b}} + {\rm{c}} + {\rm{a}} + {\rm{c}}}}{{\left( {{\rm{a}} + {\rm{c}}} \right)\left( {{\rm{b}} + {\rm{c}}} \right)}}$ = $\frac{3}{{{\rm{a}} + {\rm{b}} + {\rm{c}}}}$

Or, (a + b + 2c)(a + b + c) = 3(a + c)(b + c).

Or, a² + ab + ac + ab + b² + bc + 2ac + 2bc + 2c² = 3ab + 3ac + 3bc + 3c².

Or, a² + b² – c² = ab.

Or, $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2}}}{{2{\rm{ab}}}} = \frac{1}{2}$

Or, cosC = $\frac{1}{2}$.

So, C = 60°.

 

9) If the cosines of two of the angles of a triangle are proportional to the opposite sides, prove that the triangle is isosceles.

Solution:

Here, Let ABC be a triangle where cosA:cosB = a:b

To show: $\Delta $ABC is isosceles.

Since, cos A : cos B = a:b

Or, $\frac{{{\rm{cosA}}}}{{{\rm{cosB}}}}$ = $\frac{{\rm{a}}}{{\rm{b}}}$

Or, bcosA = acosB.

Or, b. $\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{2{\rm{bc}}}}$ = a. $\frac{{{{\rm{c}}^2} + {{\rm{a}}^2} - {{\rm{b}}^2}}}{{2{\rm{ca}}}}$

Or, b² + c² – a² = c² + a² – b².

Or, b² – a² = a² – b²

Or, 2b² = 2a²

So, a = b

So, the $\Delta $ABC is isosceles.

 

10) If $\frac{{{\rm{cosA}} + 2{\rm{cosC}}}}{{{\rm{cosA}} + 2{\rm{cosB}}}}{\rm{\:}}$ = $\frac{{{\rm{sinB}}}}{{{\rm{sinC}}}}$, prove that the triangle is either isosceles or right-angled.

Solution:

Here, $\frac{{{\rm{cosA}} + 2{\rm{cosC}}}}{{{\rm{cosA}} + 2{\rm{cosB}}}}{\rm{\:}}$ = $\frac{{{\rm{sinB}}}}{{{\rm{sinC}}}}$

Or, cosA.sinC + 2cosC.sinC = cosA.sinB + 2cosB.sinB.

Or, cosA (sinC – sinB) + sin2C – sin2B = 0

Or, cosA (sinC – sinB) + 2cos $\frac{{2{\rm{C}} + 2{\rm{B}}}}{2}$. Sin $\frac{{\left( {2{\rm{C}} - 2{\rm{B}}} \right)}}{2}$ = 0

Or, cosA.(sinC – sinB) – 2cosA.sin(C – B) = 0    [B + C = π – A]

Or, cosA {sinC – sinB – 2sin(C – B)} = 0

Or, 2sin $\frac{{{\rm{C}} - {\rm{B}}}}{2}$$\left( {\cos \frac{{{\rm{C}} + {\rm{B}}}}{2} - 2\cos \frac{{{\rm{C}} - {\rm{B}}}}{2}} \right)$ = 0

So, sin $\frac{{{\rm{C}} - {\rm{B}}}}{2}$ = 0      $\left[ {\left( {\cos \frac{{{\rm{C}} + {\rm{B}}}}{2} - 2\cos \frac{{{\rm{C}} - {\rm{B}}}}{2} \ne 0} \right)} \right]$

Or, $\frac{{{\rm{C}} - {\rm{B}}}}{2}$ = 0 à C = B

Hence, Either A = 90° or B = C.

So, the triangle is either isosceles or right angled.

 

11) If 2cosA = $\frac{{{\rm{sinB}}}}{{{\rm{sinC}}}}$ , show that the triangle is isosceles.

Solution:

Here, 2cosA = $\frac{{{\rm{sinB}}}}{{{\rm{sinC}}}}$

Or, 2.$\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{2{\rm{bc}}}}$ = $\frac{{\rm{b}}}{{2{\rm{R}}.\frac{{\rm{c}}}{{2{\rm{R}}}}}}$

Or, $\frac{{{{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}}}{{{\rm{bc}}}} = \frac{{\rm{b}}}{{\rm{c}}}$.

Or, b² + c² – a² = b²

Or, c² – a² = 0

So, c = a

So, the triangle is isosceles.

 

12) Prove that a², b², c², are in A.P. if Sin A: Sin C = Sin (A-B): Sin (B-C)

Solution:

Here, $\frac{{{\rm{sinA}}}}{{{\rm{sinC}}}}$ = $\frac{{\sin \left( {{\rm{A}} - {\rm{B}}} \right)}}{{\sin \left( {{\rm{B}} - {\rm{C}}} \right)}}$

To show: a²,b²,c² are in AP.

i.e. b² = $\frac{{{{\rm{a}}^2} + {{\rm{c}}^2}}}{2}$

Since, $\frac{{{\rm{sinA}}}}{{{\rm{sinC}}}}$ = $\frac{{\sin \left( {{\rm{A}} - {\rm{B}}} \right)}}{{\sin \left( {{\rm{B}} - {\rm{C}}} \right)}}$

Or, sinA.sin(B – C) = sinC. Sin(A – B)

Or, sin(B + C).sin(B – C) = sin (A + B).sin(A – B)                                    [A+B+C = π]

Or, sin²B – sin²C = sin²A – sin²B.

Or, $\frac{{{{\rm{b}}^2}}}{{4{{\rm{R}}^2}}}$ – $\frac{{{{\rm{c}}^2}}}{{4{{\rm{R}}^2}}}$ = $\frac{{{{\rm{a}}^2}}}{{4{{\rm{R}}^2}}}$ – $\frac{{{{\rm{b}}^2}}}{{4{{\rm{R}}^2}}}$

Or, b² – c² = a² – b²

Or, 2b² = a² + c²

So, b² = $\frac{{{{\rm{a}}^2} + {{\rm{c}}^2}}}{2}$.

 

13) In any triangle, prove that:

a) a²cotA + b²cotB + c²cotC=4∆

Solution:

L.H.S. = a²cotA + b²cotB + c²cotC

= a². $\frac{{{\rm{cosA}}}}{{{\rm{sinA}}}}$ + b²$\frac{{{\rm{cosB}}}}{{{\rm{sinB}}}}$ + c²$\frac{{{\rm{cosC}}}}{{{\rm{sinC}}}}$.

= 4R² sin²A. $\frac{{{\rm{cosA}}}}{{{\rm{sinA}}}}$ + 4R²$\frac{{{\rm{cosB}}}}{{{\rm{sinB}}}}$ + 4R² sin²C $\frac{{{\rm{cosC}}}}{{{\rm{sinC}}}}$.

= 2R² (2sinA.cosA + 2sinB.cosB + 2sinC.cosC)

= 2R² (sin2A + sin2B + sin2C)

= 2R².4sinA.sinB.sinC   [A+B+C = π]

= 8R² . $\frac{{\rm{a}}}{{2{\rm{R}}}}$, $\frac{{\rm{b}}}{{2{\rm{R}}}}$. $\frac{{\rm{c}}}{{2{\rm{R}}}}$ = $\frac{{{\rm{abc}}}}{{\rm{R}}}$ = 4∆ = R.H.S.

b) (a.sinA + b.sinB + c.sinC)²=(a² + b² + c²)(sin²A + sin²B + sin²C)

Solution:

L.H.S. = (a.sinA + b.sinB + c.sinC)²

= ${\left( {{\rm{a}}.\frac{{\rm{a}}}{{2{\rm{R}}}} + {\rm{b}}.\frac{{\rm{b}}}{{2{\rm{R}}}} + {\rm{c}}.\frac{{\rm{c}}}{{2{\rm{R}}}}} \right)^2}$

= ${\left( {\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}}}{{2{\rm{R}}}}} \right)^2}$

= (a² + b² + c²). $\left( {\frac{{{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}}}{{4{{\rm{R}}^2}}}} \right)$

= (a² + b² + c²) . $\left\{ {{{\left( {\frac{{\rm{a}}}{{2{\rm{R}}}}} \right)}^2} + {{\left( {\frac{{\rm{b}}}{{2{\rm{R}}}}} \right)}^2} + {{\left( {\frac{{\rm{c}}}{{2{\rm{R}}}}} \right)}^2}} \right\}$

= (a² + b² + c²)(sin²A + sin²B + sin²C) = R.H.S.

 

c) sinA + sinB + sinC = $\frac{{\rm{s}}}{{\rm{R}}}$

Solution:

L.H.S. = sinA + sinB + sinC

= $\frac{{\rm{a}}}{{2{\rm{R}}}}$ + $\frac{{\rm{b}}}{{2{\rm{R}}}}$ + $\frac{{\rm{c}}}{{2{\rm{R}}}}$

= $\frac{{{\rm{a}} + {\rm{b}} + {\rm{c}}}}{{2{\rm{R}}}}$ = $\frac{{2{\rm{s}}}}{{2{\rm{R}}}}$

= $\frac{{\rm{s}}}{{\rm{R}}}$ = R.H.S. 

d. acosB.cosC + bcosC.cosA + c.cosA.cosB = $\frac{\Delta }{{\rm{R}}}$

Solution:

L.H.S. = acosB.cosC + bcosC.cosA + c.cosA.cosB

= cosC (a.cosB + b.cosA) + c.cosA.cosB

= c.cosC + c.cosA.cosB    [acosB + bcosA = c]

= c.{cosC + cosA.cosB}

= c.{–cosA(A + B) + cosA.cosB}  [A + B + C = π]

= c.{–cosA.cosB + sinA.sinB + cosA.cosB}

= c.sinA.sinC

= c. $\frac{{\rm{a}}}{{2{\rm{R}}}}$. $\frac{{\rm{b}}}{{2{\rm{R}}}}$ = $\frac{{{\rm{abc}}}}{{4{\rm{R}}}}$. $\frac{1}{{\rm{R}}}$ = $\frac{\Delta }{{\rm{R}}}$ = R.H.S.

 

14) 8R² = a² + b² + c², prove that the triangle is right angled.

Solution:

Here, 8R² = a² + b² + c²

Or, 8R² = 4R² sin²A + 4R²sin²B + 4R² sin²C    [a= 2R sinA]

Or, 2 = sin²A + sin²B + sin²C.

Or, 2 = $\frac{{1 - {\rm{cos}}2{\rm{A}}}}{2}$ + $\frac{{1 - {\rm{cos}}2{\rm{B}}}}{2}$ + sin²C.

Or, 2 = $\frac{1}{2} + \frac{1}{2} - \frac{1}{2}$ (cos2A + cos2B) + sin²C.

Or, 2 = 1 – $\frac{1}{2}$.2cos (A + B).cos (A – B) + sin²C.

Or, 2 = 1 + cosC.cos(A – B) + 1 – cos²C   [A + B = π – C]

Or, 0 = cosC(cos (A – B) – cosC)

Or, 0 = cosC [cos(A – B) + cos(A + B)]

Or, 0 = cosC.2cosA.cosB

Or, 0 = cosA.cosB.cosC.

Hence, either cosA = 0 à A = 90°

Or, cos B = 0 à B = 90°.

Or, cosC = 0 à C = 90°.

So, the triangle is right angled.

 

15) If a = 10, b = 8 and c = 6, find s, $\Delta $,R and sin $\frac{{\rm{B}}}{2}$.

Solution:

Here, a = 10, b = 8 and c = 6

Now, s = $\frac{{{\rm{a}} + {\rm{b}} + {\rm{c}}}}{2}$ = $\frac{{10 + 8 + 6}}{2}$ = 12

Or, $\Delta $ = $\sqrt {{\rm{s}}\left( {{\rm{s}} - {\rm{a}}} \right)\left( {{\rm{s}} - {\rm{b}}} \right)\left( {{\rm{s}} - {\rm{c}}} \right)} $ = $\sqrt {12\left( {12 - 10} \right)\left( {12 - 8} \right)\left( {12 - 6} \right)} $

Or, $\Delta $ = $\sqrt {12.2.4.6} $ = $\sqrt {{{\left( {24} \right)}^2}} $

So, $\Delta $ = 24.

And R = $\frac{{{\rm{abc}}}}{{4\Delta }}$ = $\frac{{10.8.6}}{{4.24}}$ = 5

Also, sin $\frac{{\rm{B}}}{2}$ = $\sqrt {\frac{{\left( {{\rm{s}} - {\rm{c}}} \right)\left( {{\rm{s}} - {\rm{a}}} \right)}}{{{\rm{ca}}}}} $ = $\sqrt {\frac{{\left( {12 - {\rm{b}}} \right)\left( {12 - 10} \right)}}{{6{\rm{*}}10}}} $ = $\sqrt {\frac{{6{\rm{*}}2}}{{6{\rm{*}}10}}} $ = $\sqrt {\frac{1}{5}} $ = $\frac{1}{{\sqrt 5 }}$.

Hence, s = 12, $\Delta $ = 24, R = 5, sin $\frac{{\rm{B}}}{2}$ = $\frac{1}{{\sqrt 5 }}$