If cose A = 7÷4 prove that 1+tan2A=sec2A
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Cosec A = 7/4 => hypotenuse = 7x and perpendicular = 4x
ACC TO PYTHAGORAS THM, base^2 = 49 x^2 - 16x^2 = 33x ^2
now 1+ tan^2A = 1 + 16/33 = 49 /33
SEC^2A=HYPOTENUSE^2/BASE^2
=49/33=LHS
PROVED
Step-by-step explanation:
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