if sinx=nsin(x+2A), show that:-tan(x+A)=(1+n/1-n) tanA
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Answered by
16
Hey mate,
⚡here is your answer
⭐sin(x+2a) / sin(x) = 1/n
[sin(x+2a) + sin(x)] / [sin(x+2a) – sin(x)] = (1+n) / (1-n)
(:.Using componendo and dividendo)
2.sin(x+a).cos(a) / 2.sin(a).cos(x+a) = (1+n) / (1-n)
tan(x+a) / tan(a)
= (1+n) / (1-n)tan(x+a)
I hope it helps.
⭐mark it as brainliest.
= (1+n) × tan(a)/ (1-n)
⚡here is your answer
⭐sin(x+2a) / sin(x) = 1/n
[sin(x+2a) + sin(x)] / [sin(x+2a) – sin(x)] = (1+n) / (1-n)
(:.Using componendo and dividendo)
2.sin(x+a).cos(a) / 2.sin(a).cos(x+a) = (1+n) / (1-n)
tan(x+a) / tan(a)
= (1+n) / (1-n)tan(x+a)
I hope it helps.
⭐mark it as brainliest.
= (1+n) × tan(a)/ (1-n)
Answered by
2
Answer:
ok
Step-by-step explanation:
sinx=nsin(x+2A), show that:-tan(x+A)=(1+n/1-n) tanA
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tan(x+a) / tan(a)
= (1+n) / (1-n)tan(x+a)