Question

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

Answer 1

Question:-

If sin A + cos A = m and sec A + Cosec A = n , then n(m + 1)(m - 1) =

A) m

B) n

C) 2m

D) 2n

Answer:-

Given:

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

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

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.

Answer 2

Step-by-step explanation:

Given

• sin theta + cos theta= m, sec theta + cosec theta = n.

To Find : -

• n(m+1)(m-1)=?

Solution : -

$\sf \to (sec \: + cosec) \: ({cos \: + sin \:})^{2} - 1$

$\sf \to (sec\theta + cosec\theta)[cos^2 \theta + sin^2 \theta + 2sin\theta cos\theta - 1$

$\sf \to sec\theta \times 2sin\theta cos\theta + cosec\theta \times 2sin\theta cos\theta$

$\sf \to 2sin\theta + 2 cos\theta$

$\sf \to 2[sin\theta + cos\theta]$

= 2m

Option ( c ) 2m is Answer