Question

of
sinA=sinx +siny
_________
1- Sinx.siny
then show that,

COS A = +,- of cosx. cosy/
___________
1+sinx.siny​

Answer 1

Given : sinA = (sinx + siny)/(1 + sinx siny)

To prove : cosA = ± cosx cosy/(1 + sinx siny)

proof : sinA = (sinx + siny)/(1 + sinx siny) .....(1)

we know, sin²A + cos²A = 1

so, cosA = ± √(1 - sin²A)

from equation (1) we get,

= ± √[1 - (sinx + siny)²/(1 + sinx siny)²]

= ± √[{(1 + sinx siny)² - (sinx + siny)²}/(1 + sinx siny)²]

= ± √[(1 + sin²x sin²y + 2sinx siny - sin²x - sin²y - 2sinx siny)/(1 + sinx siny)² ]

= ± √(sin²x sin²y - sin²x - sin²y + 1)/(1 + sinx siny)

= ± √{(-sin²x(1 - sin²y) + 1(1 - sin²y)}/(1 + sinx siny)

= ± √(1 - sin²x)(1 - sin²y)/(1 + sinx siny)

= ± √(cos²x cos²y)/(1 + sinx siny)

= ± cosx cosy/(1 + sinx siny)

Therefore cosA = ± cosx cosy/(1 + sinx siny)