1.

Prove the following:   sin [(n+1)A] . sin [(n+2)A] + cos [(n+1)A] . cos [(n+2)A] = cos A 

Answer»

L.H.S. = sin [(n + 1)A] . sin [(n + 2)A] + cos [(n + 1)A] . cos [(n + 2)A] 

= cos [(n + 2)A] . cos [(n + 1)A] + sin [(n + 2)A] . sin [(n + 1)A] Let(n+2)Aa and(n+l)Ab …(i) 

∴ L.H.S. = cos a. cos b + sin a. sin b 

= cos (a — b) 

= cos [(n + 2)A — (n + I )A] 

…[From (i)] 

cos[(n+2 – n – 1)A] 

= cos A 

= R.H.S.



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