Question

if acosQ+ bsinQ=m and asinQ -bcosQ=n then a^2+b^2=​

Answer 1

Step-by-step explanation:

Given :-

acosQ+ bsinQ = m

asinQ -bcosQ = n

To find :-

Find the value of a²+b² ?

Solution :-

Given that

acosQ+ bsinQ = m -----------(1)

On squaring both sides then

=> (acosQ+ bsinQ)² = m²

This is in the form of (x+y)²

Where ,x = acosQ and y = bsinQ

we know that

(x+y)² = x²+2xy+y²

=>(acosQ)²+2(acosQ)(bsinQ)+(bsinQ)²

= m²

=> a²cos²Q+2abcosQsinQ+b²sin²Q

= m² -----(2)

and

asinQ -bcosQ=n ---------(3)

On squaring both sides then

=> (asinQ- bcosQ)² = n²

This is in the form of (x-y)²

Where ,x = asinQ and y = bcosQ

we know that

(x-y)² = x²-2xy+y²

=> (asinQ)²-2(asinQ)(bcosQ)+(bcosQ)² = n²

=> a²sin²Q-2abcosQsinQ+b²cos²Q = n² -----(4)

now

On adding (3)&(4)

a²cos²Q+2abcosQsinQ+b²sin²Q +a²sin²Q-2abcosQsinQ+b²cos²Q = m²+n²

=> a²cos²Q+b²sin²Q+a²sin²Q+b²cos²Q = m²+n²

=> a²(cos²Q+sin²Q)+b²(sin²Q+cos²Q) = m²+n²

We know that

sin²A+cos²A = 1

=> a²(1)+b²(1) = m²+n²

=> a²+b² = m²+n²

Answer:-

The value of a²+b² for the given problem is m²+n²

Used Identities:-

→ (x+y)² = x²+2xy+y²

→ (x-y)² = x²-2xy+y²

→ sin²A+cos²A = 1

Answer 2

Step-by-step explanation:

hope it helps you please mark me brainlist