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SUST-B_15-16
HSC - উচ্চতর গণিত ১ম পত্র
→
অধ্যায়-০৬ঃ ত্রিকোণমিতিক অনুপাত
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cos
θ
=
2
x
x
2
+
1
,
x
>
1
\cos \theta=\frac{2 x}{x^{2}+1}, x>1
cos
θ
=
x
2
+
1
2
x
,
x
>
1
হলে
tan
θ
+
sec
θ
=
?
\tan \theta+\sec \theta=?
tan
θ
+
sec
θ
=
?
A.
x
2
−
1
x
2
+
1
\frac{x^{2}-1}{x^{2}+1}
x
2
+
1
x
2
−
1
B.
x
2
+
1
2
x
\frac{x^{2}+1}{2 x}
2
x
x
2
+
1
C.
x
x
x
D.
2
x
2x
2
x
E.
x
2
+
1
x^{2}+1
x
2
+
1
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