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Answer 1

Question :

If neither A nor A + B is an odd multiple of π/2 and if m sinB = n sin (2A + B), then prove that :

(m + n) TanA = (m – n) Tan (A + B)

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Explaination :

Neither A nor (A – B) is an odd multiple of π/2.

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Given that, m sinB = n sin (2A + B).

$\pmb{ \frak{\dashrightarrow \: \dfrac{m}{n} = \dfrac{sin \: (2A + B)}{sinB} }}$

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Appy componendo and dividendo rule :

$\pmb{\frak{ \dashrightarrow \: \dfrac{ m+ n}{m - n} = \dfrac{sin \: (2A + B) + sinb}{sin \: (2A + B) - sinB} }}$

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$\pmb { \frak{ \dashrightarrow \: \dfrac{m + n}{m - n} = \dfrac{2sin \: \bigg( \dfrac{2A + B + B}{2} \bigg) \: cos \: \bigg (\dfrac{2A+ B-B}{2} \bigg) }{ \bigg( \dfrac{2A + B + B}{2} \bigg) \: sin \: \bigg (\dfrac{2A+ B - B}{2} \bigg) } }}$

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$\pmb{ \frak{ \dashrightarrow \: \dfrac{m + n}{m - n} = \dfrac{sin \: (A + B) \: cosA}{cos \: (A+ B) \: sinA} }}$

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$\pmb{ \frak{ \dashrightarrow \: \dfrac{m + n}{m - n} = \dfrac{tan \: (A + B)}{tan \:A } }}$

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$\pmb{ \frak{ \dashrightarrow \: (m + n) \: tanA = (m - n) \: tan \: (A+B)}}$

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Henceforth, proved!

Answer 2

Answer:

thanks..

Step-by-step explanation:

Neither A nor (A – B) is an odd multiple of π/2.

Given that, m sinB = n sin (2A + B).

Appy componendo and dividendo rule :

Henceforth, proved!