Question

If A + B + C = $\frac{\pi}{2}$ and if none of A, B, C is an odd multiple of $\frac{\pi}{2}$, then prove that
(a). cot A + cot B + cot C = cot A cot B cot C
(b). tan A tan B + tan B tan C + tan C tan A= 1 and hence show that
(c). $\sum\frac{cos(B + C)}{cos B cos C}$ = 2

Answer 1

(ii) A + B + C = π/2

tan(A + B + C) = tanπ/2

or, (tanA + tanB + tanC - tanA.tanB. tanC)/(1 - tanA.tanB - tanB.tanC - tanC.tanA) = tanπ/2

or, (tanA + tanB + tanC - tanA.tanB. tanC)/(1 - tanA.tanB - tanB.tanC - tanC.tanA) = 1/0

or, 1 - tanA.tanB - tanB.tanC - tanC.tanA = 0 ......(1)

or, 1 = tanA.tanB + tanB.tanC + tanC. tanA [hence proved]


(i) again, 1 = tanA.tanB + tanB.tanC + tanC. tanA

or, 1 = 1/cotA.cotB + 1/cotB.cotC + 1/cotC.cotA

or, 1 = cotC/cotA.cotB.cotC + cotA/cotA.cotB.cotC + cotB/cotA.cotB.cotC

or, cotA.cotB. cotC = cotA + cotB + cotC [hence proved]


(iii) $\sum\frac{cos(B + C)}{cos B cos C}$ = cos(B + C)/cosB.cosC + cos(C + A)/cosC.cosA + cos(A + B)/cosA.cosB

= (cosB.cosC - sinB.sinC)/cosB.cosC + (cosC.cosA - sinC.sinA)/cosC.cosA + (cosA.cosB - sinA.sinB)/cosA.cosB

= 1 - tanB.tanC + 1 - tanC.tanA + 1 - tanA.tanB

= 2 + (1 - tanA.tanB - tanB.tanC - tanC.tanA)

from equation (1),

= 2 + 0 = 2 = RHS

Answer 2

HELLO DEAR,



( a ) 1 = tanA.tanB + tanB.tanC + tanC. tanA

=> 1 = 1/cotA.cotB + 1/cotB.cotC + 1/cotC.cotA 

=> 1 = cotC/cotA.cotB.cotC + cotA/cotA.cotB.cotC + cotB/cotA.cotB.cotC 

=> cotA.cotB. cotC = cotA + cotB + cotC






( b ) A + B + C = π/2 

tan(A + B + C) = tanπ/2 

=> (tanA + tanB + tanC - tanA.tanB. tanC)/(1 - tanA.tanB - tanB.tanC - tanC.tanA) = tanπ/2 

=> (tanA + tanB + tanC - tanA.tanB. tanC)/(1 - tanA.tanB - tanB.tanC - tanC.tanA) = 1/0 

=> 1 - tanA.tanB - tanB.tanC - tanC.tanA = 0 --------( 1 )

=> 1 = tanA.tanB + tanB.tanC + tanC. tanA






( c ) $\bold{\sum\frac{cos(B + C)}{cos B cos C}}$ = cos(B + C)/cosB.cosC + cos(C + A)/cosC.cosA + cos(A + B)/cosA.cosB 

=> (cosB.cosC - sinB.sinC)/cosB.cosC + (cosC.cosA - sinC.sinA)/cosC.cosA + (cosA.cosB - sinA.sinB)/cosA.cosB 

=> 1 - tanB.tanC + 1 - tanC.tanA + 1 - tanA.tanB

=> 2 + (1 - tanA.tanB - tanB.tanC - tanC.tanA) 

from -------- ( 1 )

=> 2 + 0 = 2




I HOPE IT'S HELP YOU DEAR,
THANKS