Accha toh janaab humse gussa ho gye hain
if sin^2 + 3 cos^2=4 then show that tanO°=1/√3
Answers
4sin 2 θ+3(sin 2 θ+cos 2 θ)=44sin 2 θ+3=44sin 2 θ=1⇒sinθ= 21AC 2 =AB 2 +BC 24−1=AB⇒AB= 3tanθ= BABC = 31
We have:
sin
2
(
θ
)
+
3
cos
2
(
θ
)
=
4
which really is just:
sin
2
(
θ
)
+
cos
2
(
θ
)
+
2
cos
2
(
θ
)
=
4
and using the identity
sin
2
(
x
)
+
cos
2
(
x
)
=
1
for all
x
, we get:
1
+
2
cos
2
(
θ
)
=
4
(eq.A)
We carry on and simplify eq.A until we get an expression for
cos
(
θ
)
:
2
cos
2
(
θ
)
=
3
cos
2
(
θ
)
=
3
2
cos
(
θ
)
=
±
√
3
2
We also notice that
cos
2
(
θ
)
=
1
−
sin
2
(
θ
)
, so (eq.A) becomes:
1
+
2
cos
2
(
θ
)
=
1
+
2
(
1
−
sin
2
(
θ
)
)
=
4
i.e.
3
−
2
sin
2
(
θ
)
=
4
i.e.
sin
2
(
θ
)
=
−
1
2
sin
(
θ
)
=
±
√
1
2
(it should really be 'minus-plus' instead but the symbol does not exist here in Socratic).
So, now it is simply a matter of plugging these in the
tan
function:
tan
(
θ
)
=
sin
(
θ
)
cos
(
θ
)
=
±
√
1
2
±√32=±(1√3)=±√33
(it is more conventional to write it like this without the square-root in the denominator).