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If 3 Sin A + 4 cos A= 5 prove that tan A = 3/4​

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Step by step explanation:

Starting with the given equation:

3 sin A + 4 cos A = 5

We can square both sides:

(3 sin A + 4 cos A)^2 = 5^2

Expanding the left-hand side using the identity (a + b)^2 = a^2 + 2ab + b^2, we get:

9 sin^2 A + 24 sin A cos A + 16 cos^2 A = 25

Using the identity sin^2 A + cos^2 A = 1, we can replace sin^2 A with 1 - cos^2 A, giving:

9(1 - cos^2 A) + 24 sin A cos A + 16 cos^2 A = 25

Simplifying, we get:

9 - 9 cos^2 A + 16 cos^2 A + 24 sin A cos A = 25

Combining like terms, we get:

7 cos^2 A + 24 sin A cos A - 16 = 0

Dividing both sides by cos^2 A (which we assume is not equal to zero), we get:

7 + 24 tan A - 16 sec A = 0

Using the identity tan A = sin A / cos A and sec A = 1 / cos A, we can rewrite this equation as:

7 + 24 (sin A / cos A) - 16 (1 / cos A) = 0

Multiplying both sides by cos A, we get:

7 cos A + 24 sin A - 16 = 0

Now we can solve for tan A:

tan A = sin A / cos A

tan A = (3 sin A) / (4 cos A)

tan A = (3/4) (sin A / cos A)

tan A = 3/4

Therefore, we have proved that if 3 sin A + 4 cos A = 5, then tan A = 3/4.
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