Question

If cosA/cosB=n and sinA/sinB=m then (m^2-n^2)=?​

Answer 1

Answer :

Option A   $1-n^2$

Step By Step Explanation :-

$cosAcosB=n$  ,  $sinAsinB=m$

⟹ $m^2-n^2= (sinA/sinB)^2 = (cos A / cos B)^2$

⟹ $m^2-n^2= (sin^2A/sin^2B)-(cos^2A/cos^2B)$

⟹  $(m^2-n^2)sin^2B =(sin^2A-cos^2Atan^2B)$

⟹ $(m^2-n^2)sin^2B=(1-cos^2A-cos^2Atan^2B)$

⟹ $(m^2-n^2)sin^2B= [1-cos^2A(1+tan^2B)]$

⟹ $(m^2-n^2)sin^2B=(1-cos^2Asec^2B)$

⟹  $(m^2-n^2)sin^2B=(1-cos^2A/cos^2B)$

⟹ $(m^2-n^2)sin^2B=(1-n^2)$

Hence your  answer is $1-n^2$

Option A

                                                                                                                     

Answer 2

Answer :

According to question

m = Sin a/Sin b

n = Cos a/Cos b

thus putting these in 

(m^2 - n^2)Sin^2b = (Sin^2 a/Sin^2 b - Cos^2 a/Cos^2 b)Sin^2 b

taking LCM

                             = [(Sin^2 a Cos^2 b - Cos^2 a Sin^2 b)/Sin^2 b Cos^2 b]

                                      * Sin^2 b

                             = (Sin^2 a Cos^2 b - Cos^2 a Sin^2 b)/ Cos^2 b

                             = Sin^2 a - Cos^2 a Sin^2 b/Cos^2 b

                            = Sin^2 a -  Cos^2 a tan^2 b

Read more on Brainly.in - https://brainly.in/question/2182588#readmore