If cos a / cos b= m and cos a / sin b = n then show that (m2 + n2)cos2b = n2
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Answered by
48
m=cosA/cosB
n= CosA/sinB
LHS = (m^2 +n^2)cos^2B
=( cos^2A/cos^2B+cos^2A/sin^2B)×cos^2B
=cos^2A+cos^2A×cos^2B/sin^2B
= cos^2A+cos^2A×cot^2B
=cos^2A (1+cot^2B)
=cos^2A×cosec^2B
=(cosA/sinB )^2
=n^2=RHS. .....( n=cosA/sinA given ) proved
I m not getting want u did....
can u plss explain
Here given the value of among ...now Put the value of given equation in LHS side and now calculated the value u will get yr RHS
Now ur satisfied
Dat I know...but how did u solve it
Oi m not getting d way u solved it
plsss explain properly
Put the value of m and n ....and Square it and multiple that sum Cos^2 B
Answered by
33
m = cos Acos B; n = cos Asin Bnow, LHS = (m2+n2) cos2B=[cos2Acos2B + cos2Asin2B] . cos2B=[cos2A .sin2B + cos2A . cos2Bcos2B . sin2B] . cos2B=cos2A(sin2B+ cos2B)sin2B=cos2Asin2B=n2=RHS PLZZ MARKS AS BRILLANT ANSWER.
This one
PLZZ MARKS AS BRILLANT ANSWER.
Thanx
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