Math, asked by bhaveshvk18, 1 year ago

hey

50 points please solve this

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bhaveshvk18: any one know the answer

Answers

Answered by paryuljain23

I think this is useful for you.....

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bhaveshvk18: yes..but u wrote 1 - sin*2A

bhaveshvk18: ok bro..thank u

bhaveshvk18: yep...let the other answer come

Answered by siddhartharao77

Answer:

Tan A

Step-by-step explanation:

$Given:\frac{sinA-2sin^3A}{2cos^3A-cosA}$

$=\frac{sinA(1-2sin^2A)}{cosA(2cos^2A-1)}$

∴ sin²θ + cos²θ = 1.

$=\frac{sinA(sin^2A+cos^2A-2sin^2A)}{cosA(2cos^2A - (sin^2A + cos^2A))}$

$=\frac{sinA(sin^2A+cos^2A-2sin^2A)}{cosA(2cos^2A-sin^2A-cos^2A)}$

$=\frac{sinA(cos^2A-sin^2A)}{cosA(cos^2A-sin^2A)}$

$=\frac{sinA}{cosA}$

$=\boxed{TanA}$

Hope it helps!

bhaveshvk18: thanks a lot bro...for both the answers..

siddhartharao77: Thank you for brainliest!

bhaveshvk18: u derserve it bro. not need of thanks..in fact i should thank you..

bhaveshvk18: no* need

siddhartharao77: Do it in another answer also.. :-)

bhaveshvk18: done already

siddhartharao77: Thank you!

bhaveshvk18: ur wlcm bro ...hope u will help me whenever i get doubts

siddhartharao77: Sure :-)

bhaveshvk18: yep..thanks bro

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