Question #f4c39

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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