Question

please solve the following trigonometry questions

1. If sinθ=[m²-n²/m²+n²], find value of cosθ.

2.[1/sec²x+1/cosec²x] is equal to?

5 points

Answer 1

Hi ,

I'm using x instead of theta .

1 ) sinx = ( m² - n² ) / ( m² + n² )

Cosx = √ 1 - sin² x

= √ 1 - [ ( m² - n² ) / ( m² + n² ) ]²

= √[(m² + n² )²-(m² + n²)²] /√(m²+n²)²

= √ 4m²n² / ( m² + n² )

Cosx = 2mn / ( m² + n ² )

2 ) 1/sec² x + 1 / cosec² x

= Cos² x + sin² x

= 1

I hope this helps you.

:)