Question

1. If a cos 0 + b sin 0 = m and a sin 0-bcos = n, prove that
(m2 + n2) = (a + b2).
​

Answer 1

Step-by-step explanation:

Let theta represented by A.

(a cos A + b sin A)^2 = m^2

→a^2 cos^2A + b^2sin^2A + 2absinA•cosA = m^2 .....(1)

Now ,

(a sin A -b cosA)^2 = n^2

→a^2 sin^2A + b^2 cos^A -2abcosA•sinA = n^2 ....(2)

Adding (1)and (2),

a^2sin^2A + a^2cos^2A + b^2 sin^2A + b^2cos^2A = m^2 + n^2 .

a^2(sin^2A + cos^2A)+ b^2(sin^2A+ cos^2A)= m^2 + n^2

a^2 + b^2 =m^2 + n^2

Answer 2

Given that : a cos0 + b sin0 = m ...(1)

and a sin0 - b cos0 = n ... (2)

As we know that : cos0 = 1 and sin0 = 0

Then, from equation (1) and (2) :

a = m .....(3) and

- b = n => b = -n ....(4)

We have to prove that :

m² + n² = a² + b²

On taking LHS :

m² + n²

(a)² + (-b)² [ from equation (3) and (4) ]

=> a² + b² = RHS

HENCE PROVED