Question

plz help me out with this question
NO SPAMS OR I WILL REPORT YOUR ID AND I HAVE REPORTED SOMEONE'S ID ALREADY ​

Answer 1

Answer:-

(Theta is taken as "A")

Given:

sin A + Cos A = 1 -- equation (1)

On squaring both sides we get,

(Sin A + Cos A)² = 1²

Using the formula (a + b)² = a² + b² + 2ab in LHS we get,

sin² A + cos² A + 2sin A Cos A = 1

Using the identity sin² A + cos² A = 1 in LHS we get,

→ 1 + 2 sin A Cos A = 1

→ 2sin A Cos A = 1 - 1

→ 2 sin A Cos A = 0

We have to prove that,

sin A - Cos A = ± 1

LHS:

We know that,

(a - b)² = (a + b)² - 4ab

Hence,

(sin A - Cos A)² = (sin A + Cos A)² - 4sin A Cos A

Putting the values from equation (1) we get,

→ (sin A - Cos A)² = (1)² - 2*2sin A * Cos A

→ (sin A - Cos A)² = 1 - 2(0)

→ (sin A - Cos A)² = 1

→ sin A - Cos A = √1

→ sin A - Cos A = ± 1

→ LHS = RHS

Hence, Proved.