tanA+secA=x, to sidh kare ke sinA=x2+1/x2-1

1.	tanA+secA=x, to sidh kare ke sinA=x2+1/x2-1
Answer» As we know that 1 + tan²A = sec ²A sec²A - tan²A = 1 (secA + tan A) (secA - tanA) = 1 (secA - tanA) = 1/(secA + tanA) secA - tanA = 1/x ( As givensecA + tanA = x) Now adding (secA + tanA ) + ( secA - tanA) = x + 1/x 2secA = (x² + 1) /x 1/cosA = (x²+1)/2x ( As we know thatsecA = 1/cosA) cosA = 2x/(x² + 1) Cos²A = 4x²/(x² + 1)² 1 - cos²A = 1 - 4x²/(x² + 1)² sin²A = (x² + 1 )² - 4x²/(x² + 1)²(sin²A + cos²A = 1) sin²A = [(x²)² + 2x² + 1 -4x²]/(x² + 1) sin²A = [(x²)² - 2x² + 1]/(x² +1)² sinA = √(x² - 1)² / √(x² + 1)² sinA = (x² - 1)/ (x² + 1) Hence proved. true

Answer»

As we know that 1 + tan²A = sec ²A

sec²A - tan²A = 1

(secA + tan A) (secA - tanA) = 1

(secA - tanA) = 1/(secA + tanA)

secA - tanA = 1/x ( As givensecA + tanA = x)

Now adding

(secA + tanA ) + ( secA - tanA) = x + 1/x

2secA = (x² + 1) /x

1/cosA = (x²+1)/2x ( As we know thatsecA = 1/cosA)

cosA = 2x/(x² + 1)

Cos²A = 4x²/(x² + 1)²

1 - cos²A = 1 - 4x²/(x² + 1)²

sin²A = (x² + 1 )² - 4x²/(x² + 1)²(sin²A + cos²A = 1)

sin²A = [(x²)² + 2x² + 1 -4x²]/(x² + 1)

sin²A = [(x²)² - 2x² + 1]/(x² +1)²

sinA = √(x² - 1)² / √(x² + 1)²

sinA = (x² - 1)/ (x² + 1)

Hence proved.

true

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