if m=a sec A+b tan A and n=a tan A+b sec A,then prove that:m square-n square=a square-b square
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17
m²-n²=(m+n)(m-n)
={(a sec A+ b tan A)+ (a tan A+ b sec A)}{(a sec A+ b tan A)-(a tan A+ b sec A)}
={(a+b) sec A+(a+b) tan A}*{(a-b) sec A- (a-b) tan A}
=(a+b)(sec A+ tan A)*(a-b) (sec A -tan A)
= (a+b) (a-b) (sec A +tan A)(sec A- tan A)
=(a²-b²)(sec²A-tan²A)
=(a²-b²)(1+tan²A-tan²A)
=a² - b²
={(a sec A+ b tan A)+ (a tan A+ b sec A)}{(a sec A+ b tan A)-(a tan A+ b sec A)}
={(a+b) sec A+(a+b) tan A}*{(a-b) sec A- (a-b) tan A}
=(a+b)(sec A+ tan A)*(a-b) (sec A -tan A)
= (a+b) (a-b) (sec A +tan A)(sec A- tan A)
=(a²-b²)(sec²A-tan²A)
=(a²-b²)(1+tan²A-tan²A)
=a² - b²
Shrinithi:
thanks a lot
Answered by
3
LHS = tan²A - sin²A
= ( sin²A/cos²A ) - sin²A
= sin²A [ 1/cos²A - 1 ]
= sin²A( sec²A - 1 )
= sin²A tan²A
[ since , sec²A - 1 = tan²A ]
= tan²Asin²A
= RHS
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