Class X | Chapter: Trigonometric Identities
Q: If asecX + btanX + c = 0 and psecX + qtanX + r = 0, then prove that (br - qc)² - (pc - ar)² = (aq - bp)².
Note:- The question is from RD Sharma Class X, Ch.11, Page no.11.41, Example no. 42. They have done it using Cross Multiplication method studied in chapter: Linear equations in 2 variables but, I want to do it a different way. Please solve it using any other method. Thanks in advance!
Answers
Solution:-
Given:-
- a secx + b tanx + c = 0
- p secx + q tanx + r = 0
To Prove:-
- (br - qc)² - (pc - ar)² = (aq - bp)²
Proof:-
Condition |,
a secx + b tanx + c = 0
=) c = -a secx -b tanx
=) c = - ( a secx + b tanx )
Condition ||,
p secx + q tanx + r = 0
=) r = -p secx -q tanx
=) r = - ( p secx + q tanx )
L.H.S.
(br - qc)² - (pc - ar)²
Substituting the value of "c" and "r". we get,
=) [ -b ( p secx + q tanx ) + q ( a secx + b tanx ) ]² - [ -p ( a secx + b tanx ) + a ( p secx + q tanx ) ]²
=) [ - bp secx - bq tanx + aq secx + bq tanx ]² - [ - ap secx - bp tanx + ap secx + aq tanx ]²
=) [ secx ( aq - bp ) + tanx ( bq - bq ) ]² - [ secx ( ap - ap ) + tanx ( aq - bp ) ]²
=) [ secx ( aq - bp ) ]² - [ tanx ( aq - bp ) ]²
=) ( aq - bp )² sec²x - ( aq - bp )² tan²x
Taking " ( aq - bp )² " as common. We get,
=) ( aq - bp )² ( sec²x - tan²x )
=) ( aq - bp )² ( 1 ) [ ∵ sec²x -tan²x=1 ]
=) ( aq - bp )²
Hence Proved!
Answer lies in the attachment.
