9th question fast. Pls bro’s n sis’s
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Step-by-step explanation:
tanA=ntanB.........(1)
sinA=msinB.........(2)
(2)÷(1)
sinA/tanA=msinB/ntanB
cosA=(m/n)cosB
ncosA=mcosB
n²cos²A=m²cos²A........(3)
w/k, cos²A=1-sin²A
=1-m²sin²B (using eqn 2)
=1-m²(1-cos²B)
=1-m²+m²cos²B
=1-m²+n²cos²B
cos²A-n²cos²B=1-m²
cos²A(n²-1)=m²-1
cos²A=m²-1/n²-1
hence, proved
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