Question

please answer this question ​

Answer 1

Answer:

let theta be A

Step-by-step explanation:

43) given m=tan A+sinA , n=tanA-sinA

As m²-n²=4√mn

RHS

=4√mn

=4√(tanA+sinA)(tanA-sinA). {a²-b²=(a+b)( a-b)}=4√tanA²-sinA²

LHS

=m²-n²

=(tanA+sinA)²-(tanA-sinA)²

=(tanA²+sinA²+2*tanA*sinA)-(tanA²+sinA²-2*sinA*tanA)

=(tanA²-tanA²+sinA²

sinA²+2sinAtanA+2sinAtanA)

bro after opening bracket we get this

=4sinAtanA.

(as √tanA²=tanA) remember this bro Don't be confused

=4√tanA²sinA². (sinA²=1-cosA²).

=4√tanA²*(1-cosA²)

=4√tanA²-sinA²

=4√mn=RHS

hope this answer useful to you

Answer 2

Answer:

hope this answer useful to you mate