Question

cot 4x(sin 5x + sin 3x)
= 1
cot x(sin 5x - sin 3x)​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

to prove : cot 4x(sin 5x + sin 3x) = cot x(sin 5x - sin 3x)​

LHS = cot 4x(sin 5x + sin 3x)

= cot 4x [ 2 sin($\frac{5x+3x}{2}$)cos($\frac{5x-3x}{2}$) ]        

                                [ by formula sinA+sinB = 2sin($\frac{A+B}{2}$)cos($\frac{A-B}{2}$) ]

= 2 cot4x sin($\frac{8x}{2}$) cos($\frac{2x}{2}$)

= 2 $\frac{cos4x}{sin4x}$ *  sin4x * cosx

= 2  cos4x * cosx           ----------------- (1)

RHS = cot x(sin 5x - sin 3x)​

= cot x [ 2 cos($\frac{5x+3x}{2}$)sin($\frac{5x-3x}{2}$) ]        

                                [ by formula sinA-sinB = 2cos($\frac{A+B}{2}$)sin($\frac{A-B}{2}$) ]

= 2 cotx * cos($\frac{8x}{2}$) sin($\frac{2x}{2}$)

= 2 $\frac{cosx}{sinx}$ *  cos4x * sinx

= 2  cos4x * cosx           ----------------- (2)

so, we can see that (1) = (2)

=> LHS = RHS

Hence proved