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QuestionProve that 1+sin⁡x1−sin⁡x≡(tan⁡x+sec⁡x)2\frac{1+\sin x}{1-\sin x} \equiv(\tan x+\sec x)^{2}.

Studdy Solution

STEP 1

Assumptions1. We are given the trigonometric equation 1+sin⁡x1−sin⁡x≡(tan⁡x+sec⁡x)\frac{1+\sin x}{1-\sin x} \equiv(\tan x+\sec x)^{}. . We need to prove that the left-hand side (HS) is equal to the right-hand side (RHS) of the equation.
3. We will use the trigonometric identities tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x} and sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}.

STEP 2

Let's start by simplifying the right-hand side (RHS) of the equation using the trigonometric identities.
tan⁡x+sec⁡x=sin⁡xcos⁡x+1cos⁡x\tan x + \sec x = \frac{\sin x}{\cos x} + \frac{1}{\cos x}

STEP 3

Now, let's take the common denominator on the RHS.
tan⁡x+sec⁡x=sin⁡x+1cos⁡x\tan x + \sec x = \frac{\sin x +1}{\cos x}

STEP 4

Square the RHS of the equation.
(tan⁡x+sec⁡x)2=(sin⁡x+1cos⁡x)2(\tan x + \sec x)^2 = \left(\frac{\sin x +1}{\cos x}\right)^2

STEP 5

implify the RHS.
(tan⁡x+sec⁡x)2=(sin⁡x+1)2cos⁡2x(\tan x + \sec x)^2 = \frac{(\sin x +1)^2}{\cos^2 x}

STEP 6

Now, let's simplify the left-hand side (HS) of the equation using the identity 1−sin⁡2x=cos⁡2x1 - \sin^2 x = \cos^2 x.
1+sin⁡x1−sin⁡x=1+sin⁡xcos⁡2x\frac{1 + \sin x}{1 - \sin x} = \frac{1 + \sin x}{\cos^2 x}

STEP 7

Square the LHS of the equation.
(1+sin⁡x1−sin⁡x)2=(1+sin⁡xcos⁡2x)2\left(\frac{1 + \sin x}{1 - \sin x}\right)^2 = \left(\frac{1 + \sin x}{\cos^2 x}\right)^2

STEP 8

implify the LHS.
(1+sin⁡x1−sin⁡x)2=(1+sin⁡x)2cos⁡2x\left(\frac{1 + \sin x}{1 - \sin x}\right)^2 = \frac{(1 + \sin x)^2}{\cos^2 x}

STEP 9

Now, we can see that the LHS and the RHS of the equation are equal, so the original equation is true.
(+sin⁡x)2cos⁡2x≡(sin⁡x+)2cos⁡2x\frac{( + \sin x)^2}{\cos^2 x} \equiv \frac{(\sin x +)^2}{\cos^2 x}Therefore, +sin⁡x−sin⁡x≡(tan⁡x+sec⁡x)2\frac{+\sin x}{-\sin x} \equiv(\tan x+\sec x)^{2} is a true statement.

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