If sec θ + tan θ = p, prove that

(i)


(ii)


(iii)

(i) Given: sec θ + tan θ = p ……(1)

Then, (sec θ + tan θ) × = p


= p


= p


sec θ tan θ = (1/p) ……(2)


Adding equation (1) and (2), we get:


2 sec θ = p + (1/p)


sec θ =


Therefore, sec θ =


(ii) Given: sec θ + tan θ = p ……(1)


Then, (sec θ + tan θ) × = p


= p


= p


sec θ tan θ = (1/p) ……(2)


Subtracting equation (2) from (1), we get:


2tan θ = p – (1/p)


tan θ =


(iii) Since sin θ = tan θ/sec θ


=


=


=


13