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प्रश्न
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
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उत्तर
LHS = `(sinA + sinB)/(cosA + cosB) + (cosA - cosB)/(sinA - sinB) `
= `((sinA + sinB)(sinA - sinB) + (cosA + cosB)(cosA - cosB))/((cosA + cosB)(sinA - sinB))`
= `(sin^2A - sin^2B + cos^2A - cos^2B)/((cosA + cosB)(sinA - sinB))`
= `((sin^2A + cos^2A) - (sin^2B + cos^2B))/((cosA + cosB)(sinA - sinB)`
= `(1-1)/((cosA + cosB)(sinA - sinB))`
= `0/((cosA + cosB)(sinA - sinB))`
= 0
`(sinA + sinB)/(cosA + cosB) + (cosA - cosB)/(sinA - sinB) = 0`
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