plz answer this 1.....
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In∆ABC ; |_B =90°
•°•secA=AC/AB
sinC=AB/AC
tanA=BC/AB
tanC=AB/BC
sinB=sin90°=1
According to the sum
(secA×sinC-tanA×tanC)÷sinB
(1-1)÷1
(0)÷1
0
Answer is 1....
•°•secA=AC/AB
sinC=AB/AC
tanA=BC/AB
tanC=AB/BC
sinB=sin90°=1
According to the sum
(secA×sinC-tanA×tanC)÷sinB
(1-1)÷1
(0)÷1
0
Answer is 1....
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