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Question
i
f
sin
θ
+
1
2
sin
2
θ
=
m
cos
θ
এবং
sin
θ
-
1
2
sin
2
θ
=
n
cos
θ
হলে
m
2
-
n
2
=?
4√mn
4mn
1/(mn)
1/√ (mn)
ANSWER : 1
Descrption
<p style="margin-left:0px;"><strong>4√mn</strong>।</p><p style="margin-left:0px;">প্রদত্ত সমীকরণগুলি থেকে,</p><p>sin θ + 1/2 sin2 θ = m cos θ sin θ - 1/2 sin2 θ = n cos θ </p><p style="margin-left:0px;">প্রথম সমীকরণটিকে দুইবার গুণ করলে পাওয়া যায়,</p><p>sin2 θ + 1/2 sin4 θ + sin θ = 2m cos2 θ sin2 θ + 1/2 sin4 θ + sin θ = 2m(1 - sin2 θ) </p><p>sin4 θ - 2m sin2 θ + 2m + 1 = 0 </p><p style="margin-left:0px;">দ্বিতীয় সমীকরণটিকে দুইবার গুণ করলে পাওয়া যায়,</p><p>sin2 θ - 1/2 sin4 θ - sin θ = 2n cos2 θ sin2 θ - 1/2 sin4 θ - sin θ = 2n(1 - sin2 θ) </p><p>sin4 θ + 2n sin2 θ - 2n - 1 = 0 </p><p style="margin-left:0px;">এই দুটি সমীকরণের সমষ্টি নিলে পাওয়া যায়,</p><p>2 sin4 θ + 2mn = 0 </p><p>sin4 θ = -mn </p><p>sin2 θ = ± √(-mn) </p><p>sin2 θ = ± √mn </p><p style="margin-left:0px;">সুতরাং,</p><p>m2 - n2 = (sin2 θ + n2) - (sin2 θ - n2) </p><p>= 2 sin2 θ </p><p>= 2√mn </p><p style="margin-left:0px;">অতএব, উত্তর হল <strong>4√mn</strong>।</p>
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