Question

Prove LHS =RHS
cot4x ( sin5x + sin3x ) = cotx ( sin5x - sin3x)

Answer 1

ANSWER


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Answer 2

To Prove :-

• cot4x(sin5x + sin3x) = cotx(sin5x - sin3x)

Formula used :-

• sinC + sinD = 2 *sin(C+D/2) * cos(C - D/2)
• sinC - sinD = 2 *sin(C - D/2) * cos(C + D/2)
• cotA = cosA/sinA

Solution :-

Using , sinC + sinD = 2 *sin(C+D/2) * cos(C - D/2) in LHS, and, sinC - sinD = 2 *sin(C - D/2) * cos(C + D/2) in RHS , we get,

→ cot4x[2*{sin(5x+3x)/2}*{cos(5x - 3x)/2}] = cotx[2*{sin(5x - 3x)/2}*{cos(5x + 3x)/2}]

→ cot4x[ 2* sin(8x/2) * cos(2x/2)] = cotx[ 2 * sin(2x/2) * cos(8x/2) ]

→ cot4x[ 2 * sin4x * cosx] = cotx[ 2 * sinx * cos4x]

Now, using cotA = (cosA/sinA), Both sides we get,

→ (cos4x/sin4x)[2 * sin4x * cosx] = (cosx/sinx)[ 2 * sinx * cos4x]

→ cos4x * 2 * cosx = cosx * 2 * cos4x

→ 2 * cosx * cos4x = 2 * cosx * cos4x

→ LHS = RHS (Proved).