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