Question

prove that ,

(cos^2 A) / ( 2- 2sin^2 A - cosec^2 A ) =
(M^2) / (2M^2 - P^2 )


Here , M = cosA + sinA P = secA + cosecA​

Answer 1

Step-by-step explanation:

Proof :-

Given that

M = Cos A + Sin A

On squaring both sides then

=> M² = ( Cos A + Sin A )²

=> M² = Cos² A + 2 Sin A Cos A + Sin² A

Since , (a+b)² = a²+2ab+b²

=> M² = ( Sin² A + Co² A) + 2 Sin A Cos A

We know that

Sin² A + Cos² A = 1

Therefore, M² = 1 + 2 Sin A Cos A

We know that

Sin 2A = 2 Sin A Cos A

M² = 1+Sin 2A -------(1)

Given that

P = Sec A + Cosec A

On squaring both sides then

=> P² = ( Sec A + Cosec A )²

=> P² = [ (1/Cos A)² + (1/SinA)²]

Since , Sec A = 1 / Cos A and

Cosec A = 1 / Sin A

=> P² = [(Sin A+ Cos A ) /(Cos A Sin A) ]²

=> P² = (Sin A+ Cos A ) ²/ (Sin A Cos A)²

=> P² = (1+2 Sin A Cos A )/(Sin²A Cos² A)

=> P² = (1+ Sin 2A ) /(Sin² A Cos² A) ----(2)

Now,

2M²-P²

= [2(1 +Sin 2A)]-[(1+Sin 2A)/(Sin²A Cos² A)]

On taking (1+Sin 2A ) as common factor

= (1+Sin 2A) [ 2- {1/Sin² A Cos² A} ]

= (1+Sin 2A) [(2 Sin² A Cos² A-1) / Sin² A Cos² A]

Now,

RHS = M²/ (2M²-P²)

= (1+Sin 2A) / [(1+Sin 2A){(2 Sin² A Cos²A - 1) / Sin² A Cos² A } ]

On cancelling (1+Sin 2A ) in both numerator and denominator then

=> 1/ [(2 Sin² A Cos² A-1) / Sin² A Cos² A ]

=> (Sin² A Cos² A) / (2 Sin² A Cos² A - 1)

On dividing both numerator and denominator with 'Sin² A' then

=> Cos² A / [2 Cos² A - (1/ Sin² A)]

=> Cos² A / (2 Cos² A- Cosec² A )

=> Cos² A / [ 2( 1- Sin² A) - Cosec² A]

Since, Sin² A + Cos² A = 1

=> Cos² A/ (2 - 2 Sin² A - Cosec² A)

Therefore,

M²/ (2M²-P²) = Cos² A/ (2 - 2 Sin² A - Cosec² A)

Hence, Proved.

Used formulae:-

→ Sin² A + Cos² A = 1

→ (a+b)² = a²+2ab+b²

→ 1/Sin A = Cosec A

→ 1/Cos A = Sec A

→ Sin 2A = 2 Sin A Cos A

Answer 2

Step-by-step explanation:

Step-by-step explanation:

Proof :-

Given that

M = Cos A + Sin A

On squaring both sides then

=> M² = ( Cos A + Sin A )²

=> M² = Cos² A + 2 Sin A Cos A + Sin² A

Since , (a+b)² = a²+2ab+b²

=> M² = ( Sin² A + Co² A) + 2 Sin A Cos A

We know that

Sin² A + Cos² A = 1

Therefore, M² = 1 + 2 Sin A Cos A

We know that

Sin 2A = 2 Sin A Cos A

M² = 1+Sin 2A -------(1)

Given that

P = Sec A + Cosec A

On squaring both sides then

=> P² = ( Sec A + Cosec A )²

=> P² = [ (1/Cos A)² + (1/SinA)²]

Since , Sec A = 1 / Cos A and

Cosec A = 1 / Sin A

=> P² = [(Sin A+ Cos A ) /(Cos A Sin A) ]²

=> P² = (Sin A+ Cos A ) ²/ (Sin A Cos A)²

=> P² = (1+2 Sin A Cos A )/(Sin²A Cos² A)

=> P² = (1+ Sin 2A ) /(Sin² A Cos² A) ----(2)

Now,

2M²-P²

= [2(1 +Sin 2A)]-[(1+Sin 2A)/(Sin²A Cos² A)]

On taking (1+Sin 2A ) as common factor

= (1+Sin 2A) [ 2- {1/Sin² A Cos² A} ]

= (1+Sin 2A) [(2 Sin² A Cos² A-1) / Sin² A Cos² A]

Now,

RHS = M²/ (2M²-P²)

= (1+Sin 2A) / [(1+Sin 2A){(2 Sin² A Cos²A - 1) / Sin² A Cos² A } ]

On cancelling (1+Sin 2A ) in both numerator and denominator then

=> 1/ [(2 Sin² A Cos² A-1) / Sin² A Cos² A ]

=> (Sin² A Cos² A) / (2 Sin² A Cos² A - 1)

On dividing both numerator and denominator with 'Sin² A' then

=> Cos² A / [2 Cos² A - (1/ Sin² A)]

=> Cos² A / (2 Cos² A- Cosec² A )

=> Cos² A / [ 2( 1- Sin² A) - Cosec² A]

Since, Sin² A + Cos² A = 1

=> Cos² A/ (2 - 2 Sin² A - Cosec² A)

Therefore,

M²/ (2M²-P²) = Cos² A/ (2 - 2 Sin² A - Cosec² A)

Hence, Proved.

Used formulae:-

→ Sin² A + Cos² A = 1

→ (a+b)² = a²+2ab+b²

→ 1/Sin A = Cosec A

→ 1/Cos A = Sec A

→ Sin 2A = 2 Sin A Cos A

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