Question #93fd2

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Question #93fd2

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Answer 1 · 2016-10-09T08:40:53

We will use the following trigonometric identities:

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

as well as the following special products:

$a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ $a^2+2ab+b^2 = (a+b)^2$
$a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2 = (a+b)(a-b)$

$sec (2 x) + tan (2 x) = \frac{1}{cos (2 x)} + \frac{sin (2 x)}{cos (2 x)}$ $sec(2x) + tan(2x) = 1/cos(2x)+sin(2x)/cos(2x)$

$= \frac{1 + sin (2 x)}{cos (2 x)}$ $=(1+sin(2x))/cos(2x)$

$= \frac{1 + 2 sin (x) cos (x)}{{cos}^{2} (x) - {sin}^{2} (x)}$ $=(1+2sin(x)cos(x))/(cos^2(x)-sin^2(x))$

$= \frac{{cos}^{2} (x) + {sin}^{2} (x) + 2 sin (x) cos (x)}{{cos}^{2} (x) - {sin}^{2} (x)}$ $=(cos^2(x)+sin^2(x)+2sin(x)cos(x))/(cos^2(x)-sin^2(x))$

$= \frac{{(cos (x) + sin (x))}^{2}}{(cos (x) - sin (x)) (cos (x) + sin (x))}$ $=(cos(x)+sin(x))^2/((cos(x)-sin(x))(cos(x)+sin(x))$

$= \frac{cos (x) + sin (x)}{cos (x) - sin (x)}$ $=(cos(x)+sin(x))/(cos(x)-sin(x))$

$= \frac{\frac{cos (x) + sin (x)}{cos (x)}}{\frac{cos (x) - sin (x)}{cos (x)}}$ $=((cos(x)+sin(x))/cos(x))/((cos(x)-sin(x))/cos(x))$

$= \frac{1 + \frac{sin (x)}{cos (x)}}{1 - \frac{sin (x)}{cos (x)}}$ $=(1+sin(x)/cos(x))/(1-sin(x)/cos(x))$

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

Answer 2 · 2016-10-09T11:56:54

$L H S = sec 2 x + tan 2 x$ $LHS=sec2x+tan2x$

$= \frac{1}{cos (2 x)} + tan 2 x$ $=1/cos(2x)+tan2x$

$= \frac{1 + {tan}^{2} x}{1 - {tan}^{2} x} + \frac{2 tan x}{1 - {tan}^{2} x}$ $=(1+tan^2x)/(1-tan^2x)+(2tanx)/(1-tan^2x)$

$= \frac{1 + {tan}^{2} x + 2 tan x}{1 - {tan}^{2} x}$ $=(1+tan^2x+2tanx)/(1-tan^2x)$

$= \frac{{(1 + tan x)}^{2}}{(1 + tan x) (1 - tan x)}$ $=(1+tanx)^2/((1+tanx)(1-tanx))$

$= \frac{1 + tan x}{1 - tan x} = R H S$ $=(1+tanx)/(1-tanx)=RHS$

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Question #93fd2

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