Question

Hey friend please answer fast. It's really urgent. Tomorrow is my test. If tanA+ sinA=m and tanA- sinA=n, then prove that m^2-n^2=
$4 \sqrt{mn}$
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Answer 1

Answer:since, we need to prove 4√mn.

Soln,Here it goes..

Step-by-step explanation:

°•° m=tanA+sinA and n = tanA+ sin A

We consider,

LHS of given prove,

m^2-n^2=(tanA+sinA) ^2-(tanA-sinA) ^2

RHS: m^2-n^2=4tanA.sinA

Now,

mn=(tanA+sinA) (tanA-sinA)

•°•mn=tanA^2-sinA^2

mn=sin^2A(1/cos^2A-1) //cross multiply

mn=sin^2A(sin^2A/cos^2A) //from identity of trigonometry

mn=sin^2A. tan^2A

√mn=sinA.tanA

m^2-n^2=4tanA.sinA

m^2-n^2=4√mn

Hence proved.

Answer 2

Answer:

Refer in the attachment...