If sin theta + cos theta=mand sec theta+ cosec theta =n then n(m+1)(m-1)=?

sin A + cos A = m -- equation (1)

sec A + Cosec A = n -- equation (2).

value of  n(m+1)(m-1)

We have to find:

n(m + 1)(m - 1)

using (a + b)(a - b) = a² - b² we get,

⟹ n (m² - 1²)

⟹ m²n - n

Putting the values from equations (1) & (2) we get,

⟹ (sin A + cos A)² * (sec A + Cosec A) -

(sec A + Cosec A)

(a + b)² = a² + b² + 2ab

⟹ (sin² A + cos² A + 2sin A Cos A ) * (sec A + Cosec A) - sec A - Cosec A

using the identity sin² A + cos² A = 1 we get,

⟹ ( 1 + 2 sin A cos A ) * (sec A + Cosec A) - sec A - Cosec A

⟹ 1(sec A + Cosec A) + 2sin A cos A(sec A + Cosec A) - sec A - Cosec A

⟹ sec A + Cosec A + 2 sin A * cos A * sec A + 2 sin A cos A Cosec A - sec A - Cosec A

using sin A * Cosec A = 1 & cos A * sec A = 1 we get,

( sec A + Cosec A & - sec A - Cosec A are being cancelled )

⟹ 2 sin A + 2 cos A

⟹ 2 (sin A + cos A)

Putting the value sin A + cos A from equation (1) we get,

⟹ 2m

∴ n (m + 1)(m - 1) = 2m.

I hope it will help you.

Regards.

Step-by-step explanation:

m=sinθ+cosθ⟹m

2

=sin

2

θ+cos

2

θ+2sinθcosθ=1+2sinθcosθ

⟹sinθcosθ=

2

m

2

−1

n=secθ+cosecθ=

sinθcosθ

sinθ+cosθ

=

2

m

2

−1

m

⟹n(m

2

−1)=2m⟹n(m+1)(m−1)=2m

maximum area = xy = (16)(32) = 512 ft2

to maximize the area of the garden, the side of the fence parallel to the side of the house should be 32 ft long, and the other two sides should both have length 16 ft.'

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