How do you prove $(sin x + cos x) (tan x + cot x) = sec x + csc x$ $(sinx + cosx)(tanx + cotx)=secx + cscx$ ?

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How do you prove $(sin x + cos x) (tan x + cot x) = sec x + csc x$ $(sinx + cosx)(tanx + cotx)=secx + cscx$ ?

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Answer 1 · 2018-05-28T08:40:11

$(sin x + cos x) (tan x + cot x) = sec x - cos x + cos x + sin x + csc x - sin x = sec x + csc x$ $color(green)[(sinx + cosx)(tanx + cotx)=secx-cosx+cosx+sinx+cscx-sinx=secx+cscx]$

Explanation:

show below:

$(sin x + cos x) (tan x + cot x) = sec x + csc x$ $color(blue)[(sinx + cosx)(tanx + cotx)=secx + cscx]$

$L . H . S = (sin x + cos x) (tan x + cot x) =$ $L.H.S=color(blue)[(sinx + cosx)(tanx + cotx)]=$

$sin x \cdot tan x + sin x \cdot cot x + cos x \cdot tan x + cos x \cdot cot x =$ $sinx*tanx+sinx*cotx+cosx*tanx+cosx*cotx=$

$\frac{{sin}^{2} x}{cos x} + cos x + sin x + \frac{{cos}^{2} x}{sin x} =$ $sin^2x/cosx+cosx+sinx+cos^2x/sinx=$

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

$\frac{1}{cos x} - \frac{{cos}^{2} x}{cos x} + cos x + sin x + \frac{1}{sin x} - \frac{{sin}^{2} x}{sin x} =$ $1/cosx-cos^2x/cosx+cosx+sinx+1/sinx-sin^2x/sinx=$

$sec x - cos x + cos x + sin x + csc x - sin x = sec x + csc x = R . H . S$ $secx-cosx+cosx+sinx+cscx-sinx=color(blue)[secx+cscx]=R.H.S$

$Useful Trigonometric Identities$ $color(red)["Useful Trigonometric Identities"]$

${cos}^{2} θ + {sin}^{2} θ = 1$ $cos^2theta+sin^2theta=1$

$1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2theta=sec^2theta$

$sin 2 θ = 2 sin θ cos θ$ $sin2theta=2sin theta cos theta$

$cos 2 θ = {cos}^{2} θ - {sin}^{2} θ = 2 {cos}^{2} θ - 1 = 1 - 2 {sin}^{2} θ$ $cos2theta=cos^2theta-sin^2theta=2cos^2theta-1=1-2sin^2theta$

${cos}^{2} θ = \frac{1}{2} (1 + cos 2 θ)$ $cos^2theta=1/2(1+cos2theta)$

${sin}^{2} θ = \frac{1}{2} (1 - cos 2 θ)$ $sin^2theta=1/2(1-cos2theta)$

$tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$

$cot x = \frac{cos x}{sin x}$ $cotx=cosx/sinx$

$\frac{1}{cos x} = sec x$ $1/cosx=secx$

$\frac{1}{sin x} = csc x$ $1/sinx=cscx$

Answer 2 · 2018-05-28T08:49:49

Please see below.

Explanation:

We know that,

$(1) tan θ = \frac{sin θ}{cos θ} and cot θ = \frac{cos θ}{sin θ}$ $color(red)((1)tantheta=sintheta/costheta and cottheta=costheta/sintheta$

$(2) {sin}^{2} θ + {cos}^{2} θ = 1$ $color(blue)((2)sin^2theta+cos^2theta=1$

$(3) \frac{1}{sin θ} = csc θ and \frac{1}{cos θ} = sec θ$ $color(violet)((3)1/sintheta=csctheta and 1/costheta=sectheta$

We have to prove,

$(sin x + cos x) (tan x + cot x) = sec x + csc x$ $(sinx+cosx)(tanx+cotx)=secx+cscx$

We take Left Hand Side :

$LHS=(sinx+cosx)(tanx+cotx)...tocolor(red)(Apply(1)$

$LHS=(sinx+cosx)(sinx/cosx+cosx/sinx)$

$LHS=(sinx+cosx)((sin^2x+cos^2x)/(sinxcosx))$

$LHS=(sinx+cosx)(1/(sinxcosx))...tocolor(blue)(Apply(2)$

$LHS=sinx/(sinxcosx)+cosx/(sinxcosx)$

$LHS=1/cosx+1/sinx...tocolor(violet)(Apply(3)$

$LHS=secx+cscx$

$LHS=RHS$

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How do you prove $(sin x + cos x) (tan x + cot x) = sec x + csc x$ $(sinx + cosx)(tanx + cotx)=secx + cscx$ ?

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