If tan A = n tan B and sin A = m sin B, prove that cos^2 A= (m^2 -1) / (n^2 -1).
khanaffanullah:
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ya phir muh mein le
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Mera exam hai 19 th ko aur mai tumhari bakwaas sunne nahi baithi hu
ganga madar......
19 ko hai to kya itna mushkil sawaal puchega
bakwaas sunne agr nhi baitha hai to khada ho ja
aur sunn ek kaam kar............Muh mein le
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Answer:
m(cosAcosB-sinAsinB) = n(cosAcosB + sinAsinB)
(m-n)cosAcosB = (n+m)sinAsinB
(sinAsinB)/(cosAcosB) = (m-n)/(m+n)
sinA/cosA = [(m-n)/(m+n)]cosB/sinB
tanA = [(m-n)/(m+n)]cotB
QED
Step-by-step explanation:
m(cosAcosB-sinAsinB) = n(cosAcosB + sinAsinB)
(m-n)cosAcosB = (n+m)sinAsinB
(sinAsinB)/(cosAcosB) = (m-n)/(m+n)
sinA/cosA = [(m-n)/(m+n)]cosB/sinB
tanA = [(m-n)/(m+n)]cotB
QED
out of 5......2 days i was chilling...
nd 3 day i opened
oooo
4th day i started....nd 5 days....today m still studying
:-p
xD
;-|
say me some imp map pointing
there are many. .
and i haven't revised even one map
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