Answer :


Now, squaring both the sides, we get

p2 = q2 tan2θ …(1)

Now, solving LHS

Putting the value of p2 in the above equation, we get

[ 1+ tan2 θ = sec2 θ]

(from Eq. (1))

[(a + b) (a – b) = (a2 – b2)]

Now, we solve the RHS

[ 1+ tan2 θ = sec2 θ]


Hence Proved

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