Question

Prove that:cos(A+B+C)=cosA.cosB.cosC(1-tanB.tanC-tanC.tanA-tanA.tanC)​

Answer 1

Answer:

Step-by-step explanation:

To Prove :-

cos(A + B + C) = cosA.cosBcosC(1 - tanB.tanC - tanC.tanA - tanA.tanC)

Formula Required :-

cos(x + y) = cosxcosy - sinxsiny

sin(x + y) = sinxcosy + sinycosx

sinx/cosx = tanx

Solution :-

Taking L.H.S :-

= cos(A + B + C)

= cos(A + (B + C))

= cosAcos(B + C) - sinAsin(B + C)

= cosA( cosBcosC - sinBsinC) - sinA(sinBcosC + sinCcosB)

= cosAcosBcosC - cosAsinBsinC - sinAsinBcosC - sinAcosBsinC

Taking 'cosAcosBcosC' as common :-

$=cosAcosBcosC\left(1-\dfrac{sinBsinC}{cosBcosC}-\dfrac{sinAsinB}{cosAcosB}-\dfrac{sinAsinC}{cosAcosC}\right)$

$=cosAcosBcosC\left(1-\left(\dfrac{sinB}{cosB}\times\dfrac{sinC}{cosC}\right)-\left(\dfrac{sinA}{cosA}\times \dfrac{sinB}{cosB}\right)-\left(\dfrac{sinA}{cosA}\times \dfrac{sinC}{cosC}\right)\right)$

= cosAcosBcosC(1 - tanBtanC - tanAtanB - tanAtanC)

[ ∴ sinx/cosx = tanx]

= R.H.S

Hence Proved

Answer 2

To Prove :- cos(A + B + C) = cosA.cosB.cosC(1 - tanB.tanC - tanA.tanB - tanA.tanC) ?

Solution :-

taking LHS,

→ cos(A + B + C)

• taking (B + C) together,

→ cos[A + (B + C)]

• using cos(X + Y) = cos X * cos Y - sin X * sin Y

→ cos A * cos (B + C) - sin A * sin (B + C)

• again , using same formula,

→ cos A(cos B * cos C - sin B * sin C) - sin A * sin (B + C)

• using sin(X + Y) = sin X * cos Y + cos X * sin Y

→ cos A * cos B * cos C - cos A * sinB * sin C - sin A(sin B * cos C + cos B * sin C)

→ cos A * cos B * cos C - cos A * sinB * sin C - sin A * sin B * cos C - sin A * cos B * sin C

• taking cos A * cos B * cos C common now,

→ cos A * cos B * cos C[ 1 - (sin B * sin C/cos B * cos C) - (sin A * sin B/cos A * cos B) - (sin A * sin C/cos A * cos C)]

finally,

• using sin X / cos X = tan X .

→ cos A * cos B * cos C[ 1 - tan B * tan C - tan A * tan B - tan A * tan C ] = RHS (Proved) .

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