Q24 of 42 Page 193

If sec θ + tan θ = p. Prove that .

It is given


…(1)


We know that,


sec2 θ – tan2 θ = 1


⇒ sec2 θ – tan2 θ = 1 ⇒ (sec θ + tan θ)(sec θ – tan θ) = 1



…(2)


Adding (1) and (2) we get



∴


⇒


We know,



⇒


⇒


=



(∵ (a + b)2 = a2 + b2 + 2.a.b)




((∵ (a–b)2 = a2 + b2–2.a.b)



Hence Proved.


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