Question #a33f6

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Answer 1 · 2016-10-13T10:41:13

$R H S = sec 2 θ + tan 2 θ$ $RHS=sec2theta+tan2theta$

$= \frac{1 + {tan}^{2} θ}{1 - {tan}^{2} θ} + \frac{2 tan θ}{1 - {tan}^{2} θ}$ $=(1+tan^2theta)/(1-tan^2theta)+(2tantheta)/(1-tan^2theta)$

$= \frac{{(1 + tan θ)}^{2}}{(1 - tan θ) (1 + tan θ)}$ $=(1+tantheta)^2/((1-tantheta)(1+tantheta))$

$= \frac{1 + tan θ}{1 - tan θ}$ $=(1+tantheta)/(1-tantheta)$

$= \frac{tan (\frac{π}{4}) + tan θ}{1 - tan (\frac{π}{4}) tan θ}$ $=(tan(pi/4)+tantheta)/(1-tan(pi/4)tantheta)$

$= tan (\frac{π}{4} + θ) = L H S$ $=tan(pi/4+theta)=LHS$

Proved