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If the value of sec B + tan B = r, then the value of sec B - tan B is equal to: |
0 $\frac{1}{r}$ $r^2$ -r |
$\frac{1}{r}$ |
Given :- sec B + tan B = r We know , sec² B - tan² B = 1 ( sec B + tan B ) . ( sec B - tan B ) = 1 ( sec B - tan B ) = \(\frac{1 }{( sec B + tan B ) }\) As, sec B + tan B = r So , ( sec B - tan B ) = \(\frac{1 }{r }\) |
