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Question
In the given figure, AD is the median on BC from A. If AD = 8 cm and BC = 12 cm, find the value of `(1)/("sin"^2 x) - (1)/("tan"^2 x)`
Solution
Since AD is median on BC, we have
BD = DC = `(1)/(2) xx "BC" = (1)/(2) xx 12` = 6cm
ΔADB is a right-angled triangle.
∴ AB2
= AD2 + BD2
= 82 + 62
= 64 + 36
= 100
⇒ AB = 10cm
ΔADC is a right-angled triangle.
∴ AC2
= AD2 + DC2
= 82 + 62
= 64 + 36
= 100
⇒ AC = 10cm
`(1)/("sin"^2 x) - (1)/("tan"^2 x)`
= `(1)/(4/5)^2 - (1)/(4/3)^2`
= `(25)/(16) - (9)/(16)`
= `(16)/(16)`
= 1.
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