# Prove the following identities(secx sec y + tanx tan y)2 – (secx tan y + tanx sec y)2 = 1

LHS = (secx sec y + tanx tan y)2 – (secx tan y + tanx sec y)2

= [(secx sec y)2 + (tanx tan y)2 + 2 (secx sec y) (tanx tan y)] – [(secx tan y)2 + (tanx sec y)2 + 2 (secx tan y) (tanx sec y)]

= [sec2x sec2 y + tan2x tan2 y + 2 (secx sec y) (tanx tan y)] – [sec2x tan2 y + tan2x sec2 y + 2 (sec2x tan2 y) (tanx sec y)]

= sec2x sec2 y - sec2x tan2 y + tan2x tan2 y - tan2x sec2 y

= sec2x (sec2 y - tan2 y) + tan2x (tan2 y - sec2 y)

= sec2x (sec2 y - tan2 y) - tan2x (sec2 y - tan2 y)

We know that sec2x – tan2x = 1.

= sec2x × 1 – tan2x × 1

= sec2x – tan2x

= 1

= RHS

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
Conditional Identities31 mins
Trigonometry ratios of allied angles | Interactive Quiz38 mins
Trigonometric Functions - 0568 mins
Trigonometric Functions - 0152 mins
Graphs of trigonometric ratios | Trigonometric Identities39 mins
Trigo ratios for compound angles0 mins
Quiz on Graph of Trigonometric Ratios40 mins
Trigonometric Functions - 0366 mins
Trigonometric Functions - 0658 mins
Trigonometric Series45 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
view all courses
RELATED QUESTIONS :

Prove that :

where

RD Sharma - Mathematics

The value of is

Mathematics - Exemplar

Prove that

(i)

(ii)

(iii)

RS Aggarwal - Mathematics

If then tan (2A + B) is equal to

Mathematics - Exemplar