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NDC_2020
HSC - উচ্চতর গণিত ১ম পত্র
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অধ্যায়-০৭ঃ সংযুক্ত ও যৌগিক কোণের ত্রিকোণমিতিক অনুপাত
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f
(
x
)
=
cos
x
f(x)=\cos x
f
(
x
)
=
cos
x
ক. প্রমাণ কর যে,
tan
4
5
∘
+
θ
2
tan
4
5
∘
−
θ
2
=
2
cos
θ
−
1
2
cos
θ
+
1
\tan \frac{45^{\circ}+\theta}{2} \tan \frac{45^{\circ}-\theta}{2}=\frac{\sqrt{2} \cos \theta-1}{\sqrt{2} \cos \theta+1}
tan
2
4
5
∘
+
θ
tan
2
4
5
∘
−
θ
=
2
c
o
s
θ
+
1
2
c
o
s
θ
−
1
খ. প্রমাণ কর যে,
{
f
(
x
)
}
3
+
{
f
(
12
0
∘
+
x
)
}
3
+
{
f
(
24
0
∘
+
x
)
}
3
=
3
4
f
(
3
x
)
\{f(x)\}^{3}+\left\{f\left(120^{\circ}+x\right)\right\}^{3}+\left\{f\left(240^{\circ}+x\right)\right\}^{3}=\frac{3}{4} f(3 x)
{
f
(
x
)
}
3
+
{
f
(
12
0
∘
+
x
)
}
3
+
{
f
(
24
0
∘
+
x
)
}
3
=
4
3
f
(
3
x
)
গ.
A
+
B
+
C
=
π
2
\mathrm{A}+\mathrm{B}+\mathrm{C}=\frac{\pi}{2}
A
+
B
+
C
=
2
π
হলে, প্রমাণ কর যে,
{
(
A
)
}
2
+
{
f
(
B
)
}
2
−
{
f
(
C
)
}
2
−
2
f
(
A
)
f
(
B
)
}
(
π
2
−
C
)
=
0
\left.\{(\mathrm{A})\}^{2}+\{f(\mathrm{~B})\}^{2}-\{f(\mathrm{C})\}^{2}-2 f(\mathrm{~A}) f(\mathrm{~B})\right\}\left(\frac{\pi}{2}-\mathrm{C}\right)=0
{(
A
)
}
2
+
{
f
(
B
)
}
2
−
{
f
(
C
)
}
2
−
2
f
(
A
)
f
(
B
)
}
(
2
π
−
C
)
=
0
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