If
cot
x
=
1
√
5
, find the value of
cosec
2
x
+
sec
2
x
cosec
2
x
−
sec
2
x
.
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Answered by
1
Answer:
let given ΔABC
BD = 3, AC = 12, AD = 4
In right angled ΔABD
By Pythagoras theorem
AB
2
=AD
2
+BD
2
AB
2
=(4)
2
+(3)
2
AB
2
=16+9
AB
2
=25
AB
2
=(5)
2
AB = 5
In right angled triangle ACD
By Pythagoras theorem,
AC
2
=AD
2
+CD
2
CD
2
=AC
2
−AD
2
CD
2
=(12)
2
−(4)
2
CD
2
=144−16
CD
2
=128
CD=√128
CD=√64×2CD
=8√2
(i) sin x = perpendicular/Hypotenuse
= AD/AB
= 4/5
(ii) cot x = Base/Perpendicular
= BD/AD
=3/4
(iii) cot x = Base/ Perpendicular
BD/AD
= 3/4
cosec x = Hypotenuse / Perpendicular
AB/BD
= 5/4
cot
2
x−cosec
2
x
=(3/4)
2
−(5/4)
2
= 9/16 – 25/16
(9 -25)/16
= -16/16
= -1
Step-by-step explanation:
Hope it will help you
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