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# (*)If tan b =( n tan a ) / 1 + (1-n) tan square a, then tan (a-b) =

21 Points
4 years ago
tanA-B= tanA-tanB                      { by substituting the value of tanB we get }
1+tanAtanB

=tanA -    NtanA
1+(1-N)tan2A       {by cross multiplication we get }
1 -   tanA × NtanA
1+(1-N)tan2a

= tanA+tan3A-Ntan3A-NtanA
1+(1-N)tan2A
1+tan2A-Ntan2A+Ntan2A
1+(1-N)tan2A

= tanA+tan3A-Ntan3A-NtanA   ×          1+(1-N)tanA
1+(1-N)tanA                            1+tan2A

=  tanA-NtanA+tan3A-Ntan3A    …...{ now by taking tanA and N as comman we get }
1+tan2A

=  (1-N)tanA + (1-N)tan3   …..{ now by taking (1-N)tanA as comman we get }
1+tan2A

= (1-N)tanA A }2{ 1+tan
​A21+tan

Ans =    (1-N)tanA