Prove that :- {(1+cosA)^2 +(1-cosA) ^2} ÷ {sin^2A} = 2 (1+2cot^2A)

Step-by-step explanation:

(a + b)² + (a - b)² = 2(a² + b²)

a = 1 ; b = Cos A

(1 + Cos A)² + (1 - Cos A)² = 2(1² + Cos² A)

                                         = 2(1 +Cos² A)   ----------(I)

LHS = { (1 + Cos A)² + (1 - Cos A)²} ÷ { Sin² A}

      = { 2( 1 + Cos² A )} ÷ Sin² A                  {From (I)}

      = { 2 (Sin² A + Cos² A + Cos² A) } ÷ { Sin² A}    {Here, Sin² A + Cos² A = 1 }

     = 2 (Sin² A + 2Cos² A) }  ÷ { Sin² A}    

      = 2( 1 + 2 Cot²  A) =RHS

the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

spinner 1 spinner 2

1                     1

1                     2

1                     3

1                     4

2                   1

2                   2

2                   3

2                   4   = not 1, and 4

3                   1

3                   2

3                   3

3                   4 = not 1, and 4

only 2 outcomes have passed the given condition.

step-by-step explanation:

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