Answer :

Given: a sec θ + b tan θ = x …….(1)


a tan θ + b sec θ = y …….(2)


Square equation (1) and (2) on both sides:


a2 sec2 θ + b2 tan2 θ + 2ab sec θ tan θ = x2 …….(3)


a2 tan2 θ + b2 sec2 θ + 2ab sec θ tan θ = y2 ……..(4)


Subtract equation (4) from (3):


[a2 sec2 θ + b2 tan2 θ + 2ab sec θ tan θ] – [a2 tan2 θ + b2 sec2 θ + 2ab sec θ tan θ] = x2 – y2


a2 (sec2 θ – tan2 θ) + b2 (tan2 θ – sec2 θ) = x2 – y2


a2 – b2 = x2 – y2 (sec2 θ = 1 + tan2 θ)


Hence, proved.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Trigonometric IdentitiesTrigonometric IdentitiesTrigonometric Identities39 mins
Trigonometric Identities-IITrigonometric Identities-IITrigonometric Identities-II43 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses