Question

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

Answer 1

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

Answer 2

Given:

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

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

Find:

value of n(m+1)(m-1)

Solution:

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.