Question

OR
If cosec 6 - sin 0 = m³ and sec 0 - cos 0 = n³ then prove that m⁴n² + m²n⁴ = 1.​

Answer 1

Answer:

Let theta=x

cosecx-sinx=m^3

1/sinx -sinx =m^3

(1-sin^2x)/sinx=m^3

cos^2x/sinx=m^3……………(1)

secx-cosx=n^3

1/cosx-cosx=n^3

(1-cos^2x)/cosx=n^3

sin^2x/cosx=n^3………………..(2)

Multiply eq.(1) & (2).

(cos^2x/sinx)×((sin^2x/cosx)=m^3×n^3

cosx.sinx=m^3.n^3………………….(3)

Divide eq.(1)by (2)

(cos^2x/sinx)÷(sin^2x/cosx)=m^3÷n^3

cos^3x/sin^3x=m^3/n^3

cosx/sinx=m/n………………………..(4)

Multiply eq.(3) & (4)

cosx.sinx×cosx/sinx=m^3.n^3×m/n

cos^2x=m^4.n^2………………………….(5)

Divide eq.(3)by (4)

cosx.sinx÷(cosx/sinx)=m^3.n^3÷m/n

sin^2x=m^2.n^4……………………..(6)

On adding eq.(5) & (6)

cos^2x+sin^2x=m^4.n^2+m^2.n^4.

1 = m^4.n^2+ m^2.n^4 , proved.

Answer 2

Answer:

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