Question

If SinA +SinB = 2 then find out sin²A + Sin²B.

Answer 1

Step-by-step explanation:

We all know that sin is cyclic function from R->[0,1].

So, possible solutions for

sinA +sinB =2

This can only be possible when both sin functions are in maximum.

Hence,

SinA = 1

and A= n (π/2). Where n is odd

And similarly,

B= n (π/2). Where n is odd

Therefore

A+B = (m+n)(π/2) where m,n are odd

A+B = k(π/2) and k is even (as the sum of two odd numbers is even.)

As Sin(A+B)= 0, For all even multiples of π/2

Hence,

Sin(A+B)=0

Hope it will help u ....

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Answer 2

Answer:

0 is your correct answer......