# If cosec A =7/4,prove that 1+tanÂ²A=secÂ²A

Given that,

cosec a = 7 /4 = hypotenuse / perpendicular

hypotenuse = 7 and perpendicular = 4

therefore,

base = √( hypotenuse)² - (perpendicular)²)

base = √7² - 4²

base = √33

therefore,

tan a = p/b = 4/√33 and sec a = h/b = 7/√33

therefore,
1 + tan²a = 1 + ( 4/√33)² = 1 + 16 / 33 = ( 33 + 16 /33) = 49/33.
and,
sec²a = ( 7/√33)² = 49 / 33.
hence,  1 + tan²a = sec²a

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