Question

If a sin^2theta+b cos^2theta=sin theta ×cos theta & a sin theta- b cos theta=0,then prove that a^2+b^2=1

Answer 1

From the formula for
sin
(
2
θ
)
we have

sin
2
θ
=
2
sin
θ
cos
θ

From Pythagoras theorem we have

sin
2
θ
+
cos
2
θ
=
1

So

sin
2
θ
sin
θ

=
(
2
sin
θ
cos
θ
)
sin
θ

=
2
sin
2
θ
cos
θ

=
2
(
1
−
cos
2
θ
)
cos
θ

Putting this together with your equation, we get

2
(
1
−
cos
2
θ
)
cos
θ
=
cos
θ

If
cos
θ
=
0
then both sides will be zero.
So some solutions to the original problem are:

θ
=
π
2
+
n
π
for all
n
in
Z
.

On the other hand, if
cos
θ
≠
0
, divide both sides of the equation by
cos
θ
to get

2
(
1
−
cos
2
θ
)
=
1

Divide both sides by 2 to get

1
−
cos
2
θ
=
1
2

So
cos
2
θ
=
1
2
and
cos
θ
=
±
1
√
2

This is true for

θ
=
π
4
+
n
π
2
for all
n
in
Z
.
I think this will help.