Question #f4c39

Question

952 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-20T06:48:01

Given #tanP/tanQ=1/5=>tanQ=5tanP#

Now #LHS=cot(P+Q)=1/tan(P+Q)#

#=(1-tanPtanQ)/(tanP+tanQ)#

#=(1-5tanPxxtanP)/(tanP+5tanP)#

#=(1-5tan^2P)/(6tanP)#

#=((1-5tan^2P)xxcos^2P)/(6tanPxxcos^2P)#

#=(cos^2P-5sin^2P)/(6sinPxxcosP)#

#=(6cos^2P-5cos^2P-5sin^2P)/(3xxsinPxxcosP)#

#=(6cos^2P-5(cos^2P+sin^2P))/(3xxsin2P)#

#=1/3csc(2P)(6cos^2P-5)=RHS#

Proved