Math, asked by Murali135, 10 months ago

sin(n+1)A-sin(n-1)A/cos(n+1)A+2cosnA+cos(n-1)A

Answers

Answered by vikykumar1204

0

hence answer is tanA

sin(n+1)A-sin(n-1)A/cos(n+1)A+2cosA+cos(n-1)A

= sin nA +sinA-sin nA+sinA/cosnA+cosA+2cosA+cosnnA-cosA

=2sinA/2cosA

=tanA

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