Question

sinA+cosA/sinA-cosA + sinA-cosA/SinA+cosA=1/sin^2A-cos^2=2/1-2cos^2A=2sec^2A/tan^2A-1

Answer 1

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Mark BRAINLIEST thank u

Answer 2

$i) \frac{sinA+cosA}{sinA-cosA} +\frac{sinA-cosA}{sinA+cosA} \\= \frac{(sinA+cosA)^{2} + (sinA-cosA)^{2}}{(sinA-cosA)(sinA+cosA)}\\= \frac{2(sin^{2}A+cos^{2}A)}{sin^{2}A-cos^{2}A}$

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We know that,

• 1) (a+b)² + (a-b)² = 2(a²+b²)
• 2)(a-b)(a+b) = a² - b²
• 3) sin²A + cos²A = 1

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$= \frac{2}{sin^{2} A - cos^{2} A }\: ---(1)$

$ii ) \frac{2}{sin^{2} A - cos^{2} A }\: [ From \:(1) ]$

$= \frac{2}{1-cos^{2} A - cos^{2} A } \\= \frac{2}{1- 2cos^{2} A }\: --(2)$

$iii ) \frac{2}{sin^{2} A - cos^{2} A } \: [ From \:(1) ]$

/* Dividing numerator and denominator by cos² A , we get */

$= \frac{ \frac{2}{cos^{2} A}}{\frac{sin^{2}A}{cos^{2} A} - \frac{cos^{2}A}{cos^{2}A}}$

$= \frac{2sec^{2} A}{ tan^{2} A - 1 } \: --(3)$

$Hence \:proved$

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