(sinA -2sin^2A)/ (2cos^3A-cosA)=tanA.
Adarsh9450668057:
2SIN^3 A HAI.2SIN^2A KE JAGAH PAR,
issi question main??
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Answered by
3
(sinA - 2sin^3A)/ (2cos^3A - cosA)
sinA(1 - 2sin^2A) / cosA (2cos^2A - 1)
sinA {1 - 2 (1 - cos^2A) } / cosA (2cos^2A - 1) (sin^2A = 1 - cos^2A)
sinA(1 - 2+2cos^2A) / cosA (2cos^2A - 1)
sinA (2cos^2A - 1) / cosA (2cos^2A - 1)
sinA/ cosA {(2cos^2A - 1) / (2cos^2A - 1) } cancel each other
tanA
you have wrong question...there is sin^3A instead of sin^2A
hope it's clear to you
sinA(1 - 2sin^2A) / cosA (2cos^2A - 1)
sinA {1 - 2 (1 - cos^2A) } / cosA (2cos^2A - 1) (sin^2A = 1 - cos^2A)
sinA(1 - 2+2cos^2A) / cosA (2cos^2A - 1)
sinA (2cos^2A - 1) / cosA (2cos^2A - 1)
sinA/ cosA {(2cos^2A - 1) / (2cos^2A - 1) } cancel each other
tanA
you have wrong question...there is sin^3A instead of sin^2A
hope it's clear to you
ek sath sahi nhi rhta
hey @adharsh
is it sin^2 A or sin^3A
SIN ^3 A
bhai dubara post kar de alag karke questions ko
humare liye bhi easy rahega aur tumhare liye bhi
TUMNE BHI TO SIN^3 A LIYA HAI.
AUR MERE PASS ITNE POINTS NHI HAI KI MAY DUBARA SE POST KARU.
oh..
HA
Answered by
4
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