Question

If tan A+sin A = m and tan A - sin A = n;

prove that $m^2-n^2=4\sqrt{mn}$

Answer 1

Answer:

Step-by-step explanation:

Hey mate here is your answer:

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Question:

→$m^2-n^2=(m+n)(m-n),$

→($tanA+sinA+tanA-sinA$)($tanA+sinA-tanA+sinA$),

→($2\:tanA$)($2\:sinA$),

→$4\:tanA\:sinA$.                                  ...(I)

→$4\sqrt{mn}=4\sqrt{(tanA+sinA)(tanA-sinA)} ,$

→$4\sqrt{tan^2A-sin^2A}$,

→$4\sqrt{\frac{sin^2A}{cos^2A}-sin^2A } ,$

→$4sin\:A\sqrt{sec^2A-1} ,$                             [∵$\frac{1}{cos^2A} =sec^2A$],

→$4 sinA.tanA$  ,                        ...(II)    [∵$sec^2A-1=tan^2A$],

→$m^2-n^2=4\sqrt{mn}.$                              [From (I) and (II)].

                           Hence Proved.

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Hope it helps you.

Answer 2

Step-by-step explanation:

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Thanks for the free points:)

Btw I don't know the answer....hehehe