in triangle ABC, 4CosACosB is equal to

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in triangle ABC, 4CosACosB is equal to

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kumarsarvksha1 13th Jul, 2020

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Nikhitha Sai 22nd Aug, 2020

In a triangle ABC cos(A+B)= cosAcosB-sinAsinB. cos(A-B)= cosAcosB+sinAsinB.

there by, by adding these two and multiplying the result by 2 times you get the desired answer. 4cosAcosB = 2[cos(A+B) + cos (A-B)]. This can also be done by interchanging the signs and the answer will remain the same.

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