Question

a cos theta + b sin theta = m
and a sin theta + b cos theta = n
Than find a sq. + b sq. =?​

Answer 1

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a cos theta + b sin theta = m

and a sin theta + b cos theta = n

Than find a sq. + b sq. =?

$\huge{\tt{\colorbox{pink}{AnswEr:}}}$

And the given is proved, m²+ n²= a²+ b²

Step-by-step explanation:

Given:

m = a cos theta - b sin theta 

Solution:

Therefore,

m² = (a cos theta - b sin theta)²

m²= (a² Cos²theta) + (b² Sin²theta) - (2ab Cos theta Sin theta)   ------------ 1

Also given ,

n = a sin theta + b cos theta 

Therefore,

n² = (a sin theta + b costheta)²

n²= (a²Sin²theta) + (b²Cos²theta) + (2ab Cos theta Sin theta)   ------------ 2

Adding  1 and 2

m²+ n² = (a²Cos²theta) + (b² Sin²theta) - (2ab Cos theta Sin theta)   +  (a² Sin²theta) + (b² Cos²theta) + (2ab Cos theta Sin theta)

then, Cancelling,

(2ab Cos theta Sin theta) and - (2ab Cos theta Sin theta)

m² + n² = (a²Cos²theta) + (b² Sin²theta) +

(a² Sin²theta) + (b² Cos²theta)

Bringing a² terms and b2 terms together,

m² + n² = (a² Cos²theta) + (a²Sin²theta) + (b²Sin²theta) + (b²Cos²theta)

m² + n² = a³ (Cos²theta + Sin²theta) + b²(Sin²theta + Cos²theta)

By the identity Sin²theta + Cos²theta = 1

m² + n² = a²(1) + b² (1)

m²+ n²= a²+ b²

Hence proved.