Question

is there any genius who will tell me the formula of sin (a+b+c) and cos (a+b+c) and tan (a+b+c)​

Answer 1

Answer:

EXAMPLE

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) = 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 = LHS