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If tanA.tanB=1/2,t h e n(5-3cos2A)(5-3co...

If `tanA.tanB=1/2,t h e n(5-3cos2A)(5-3cos2B)=` (a)2 (b) 8 (c) 12 (d) 16

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`tanA*tanB = 1/2`
`=>tanA = 1/(2tanB)`
`=>tan^2A = 1/(4tan^2B)`
Now, `(5-3cos2A)(5-3cos2B)`
`=(5-3((1-tan^2A)/(1+tan^2A)))(5-3((1-tan^2B)/(1+tan^2B)))`
`=((5+5tan^2A-3+3tan^2A)/(1+tan^2A))((5+5tan^2B-3+3tan^2B)/(1+tan^2B))`
`=((2+8tan^2A)/(1+tan^2A))((2+8tan^2B)/(1+tan^2B))`
`=4*((1+4tan^2A)/(1+tan^2A))((1+4tan^2B)/(1+tan^2B))`
...
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