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)`

Answer 1

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.