Trig Identities
prove
sinx+1 \ 1-sinx = (tanx+secx)^2
If your expression mean:
( sin x + 1 ) / ( 1 - sin x ) = ( tan x + sec x )²
then:
( 1 + sin x ) / ( 1 - sin x ) = ( tan x + sec x )²
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Remark:
In expression:
( 1 + sin x ) / ( 1 - sin x )
Multiply numerator and denominator by 1 + sin x
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( 1 + sin x ) ∙ ( 1 + sin x ) / ( 1 - sin x ) ∙ ( 1 + sin x ) = tan²x + 2 ∙ tan x ∙ sec x + sec²x
( 1 + sin x )² / ( 1 ∙ 1 - sin x ∙ 1 + sin x ∙ 1 - sin x ∙ sin x ) = tan²x + 2 tan x ∙ sec x + sec²x
( 1 + sin x )² / ( 1 - sin x + sin x - sin²x ) = tan²x + 2 tan x ∙ sec x + sec²x
( 1 + sin x )² / ( 1 - sin²x ) = tan²x + 2 tan x ∙ sec x + sec²x
( 1² + 2 ∙ 1 ∙ sin x + sin²x ) / cos²x = tan²x + 2 tan x ∙ sec x + sec²x
( 1 + 2 sin x + sin²x ) / cos²x = tan²x + 2 tan x ∙ sec x + sec²x
1 / cos²x + 2 sin x / cos²x + sin²x / cos²x = tan²x + 2 tan x ∙ sec x + sec²x
sec²x + 2 sin x / ( cos x ∙ cosx )+ tan²x = tan²x + 2 tan x ∙ sec x + sec²x
sec²x + 2 sin x ∙ 1 / ( cos x ∙ cosx )+ tan²x = tan²x + 2 tan x ∙ sec x + sec²x
sec²x + 2 ( sin x / cos x ) ∙ 1 / cosx + tan²x = tan²x + 2 tan x ∙ sec x + sec²x
sec²x + 2 tan x ∙ sec x + tan²x = tan²x + 2 tan x ∙ sec x + sec²x
tan²x + 2 tan x ∙ sec x + sec²x = tan²x + 2 tan x ∙ sec x + sec²x
I would do it this way:
LS = (sinx+1)/(1-sinx)
= (sinx+1)/(1-sinx) * (sinx+1)/(1+sinx)
= (sinx+1)^2 /(1 - sin^2 x)
= (sinx + 1 )^2 / ((1+sinx)(1-sinx))
= (1+sinx)/(1-sinx)
RS = ( tan x + sec x )²
= (sinx/cosx + 1/cosx)^2
= ( (sinx+1)/cosx)^2
= (sinx+1)^2/cos^2 x
= (sinx+1)^2/(1- sin^2x) = (sinx+1)(sinx+1)/((1-sinx)(1+sinx)
= (sinx+1)/(1-sinx)
= LS
To prove the given trigonometric identity, we will start with the left side of the equation and simplify it step by step until it matches the right side of the equation.
Left side: (sin(x) + 1) / (1 - sin(x))
Step 1: To simplify this expression, we will multiply both the numerator and the denominator by the conjugate of the denominator, which is (1 + sin(x)). This will help us eliminate the fraction:
[(sin(x) + 1) / (1 - sin(x))] * [(1 + sin(x)) / (1 + sin(x))]
Step 2: Now, let's simplify the numerator by distributing the terms:
= [(sin(x) + 1)(1 + sin(x))] / (1 - sin(x))
= (sin^2(x) + sin(x) + sin(x) + 1) / (1 - sin(x))
= (sin^2(x) + 2sin(x) + 1) / (1 - sin(x))
Step 3: Next, let's simplify the denominator by using the difference of squares formula:
= (sin^2(x) + 2sin(x) + 1) / [(1 + sin(x))(1 - sin(x))]
= (sin^2(x) + 2sin(x) + 1) / (1 - sin^2(x)) [Since (1 + sin(x))(1 - sin(x)) = 1 - sin^2(x)]
Step 4: Rearrange the numerator to match the right side of the equation:
= [(1 + sin(x))^2] / (1 - sin^2(x))
= [(1 + sin(x))^2] / [(1 - sin(x))(1 + sin(x))] [Using the difference of squares: 1 - sin^2(x) = (1 - sin(x))(1 + sin(x))]
= [(1 + sin(x))^2] / [(cos(x))(sec(x))] [Using sin^2(x) = 1 - cos^2(x)]
= [(1 + sin(x))^2] / [cos(x) sec(x)]
Step 5: Recall that sec(x) is equal to 1/cos(x):
= [(1 + sin(x))^2] / (cos(x) * (1 / cos(x)))
= [(1 + sin(x))^2] / 1
= (1 + sin(x))^2
Right side: (tan(x) + sec(x))^2
Since the left side (1 + sin(x))^2 matches the right side (tan(x) + sec(x))^2, we have proven the given trigonometric identity.
Therefore, (sin(x) + 1) / (1 - sin(x)) = (tan(x) + sec(x))^2.