In the adjoining figure, Find : sin , cos O & tan 0. Also find : sin a, cos a and tan a. B In the adjoining figure, 2O = 90° & - D с =
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Answer:
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Step-by-step explanation:
AB = a
⇒ AD + DB = a
⇒ AD + AD = a
⇒ 2 AD = a
⇒ AD =
2
a
Thus, AD = DB =
2
a
By Pythagoras theorem, we have
AC
2
=AB
2
+BC
2
⇒b
2
=a
2
+BC
2
⇒BC
2
=b
2
−a
2
⇒BC=
b
2
−a
2
Thus, in ΔBCD, we have
Base = BC =
b
2
−a
2
and Perpendicular = BD =
2
a
Applying Pythagoras theorem in Δ BCD, we have
⇒BC
2
+BD
2
=CD
2
⇒(
b
2
−a
2
)
+(
2
a
)
2
=CD
2
⇒CD
2
=b
2
−a
2
+
4
a
2
⇒CD
2
=
4
4b
2
−4a
2
+a
2
⇒CD
2
=
4
4b
2
−3a
2
⇒CD=
2
4b
2
−3a
2
Now,
sin
2
θ+cos
2
θ=(
4b
2
−3a
2
a
)
2
+(
4b
2
−3a
2
2
b
2
−a
2
)
2
⇒sin
2
θ+cos
2
θ=
4b
2
−3a
2
a
2
+
4b
2
−3a
2
4(b
2
−a
2
)
⇒sin
2
θ+cos
2
θ=
4b
2
−3a
2
4b
2
−3a
2
=1.
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