Question

Hey maths aryabhattas !! Plz tell me what is the formula of cos(A+B)/cos(A-B) ....... answer kardo yarr..​

Answer 1

Question :- Simplify cos(A+B)/cos(A-B) .

→ cos(A + B) = cosA*cosB - sinA*sinB

→ cos(A - B) = cosA*cosB + sinA*sinB

we can Learn Them as ,

→ cos(A ± B) = cosA*cosB ∓ sinA*sinB

Now, Putting Both Formula we get :-

→ cos(A + B) / cos(A - B)

→ [ cosA*cosB - sinA*sinB ] / [ cosA*cosB + sinA*sinB ]

Now, Divide Both Numerator & Denominator by cosA*cosB , we get,

→ [ { cosA*cosB - sinA*sinB } / ( cosA *cosB ) ] / [ { cosA*cosB + sinA*sinB } / ( cosA *cosB ) ]

→ [ (cosA*cosB/cosA*cosB) - (sinA*sinB/cosA*cosB) ] / [ { (cosA*cosB/cosA*cosB) + (sinA*sinB/cosA*cosB) ]

→ [ { 1 - (sinA*sinB/cosA*cosB) ] / [ { 1 + (sinA*sinB/cosA*cosB) ]

Now, using sinA/cosA = tanA , we get,

→ [ 1 - tanA*tanB ] / [ 1 + tanA*tanB ] . (Ans).

Answer 2

$\huge\tt{SOLUTION:}$

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$\huge\tt{SIMPLIFYING:}$

↪cos(A+B) = cosA × cosB-sinA×sinB

↪cos(A-B) = cosA×cosB + sinA ×sinB

↪cos(A±B) = cosA×cosB ±sinA×sinB

↪cos(A+B) = cosA×cosB ± sinA×sinB

↪cos(A+B)/cos(A-B)

↪[cosA×cosB-sinA×sinB]/[cosA×cosB+sinA×sinB]

↪[{cosA×cosB-sinA×sinB}/{cosA×cosB}]/[{cosA×cosB)]

↪[(cosA×cosB/cosA×cosB)-(sinA×sinB/cosA×cosB)]/[{1-(sinA×sinB/cosA×cosB)]}/[{1+(sinA×sinB/cosA×cosB)}]

↪[1-tanA×tanB]/[1+tanA×tanB]

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