C-09-Q111mediumsingle_mcq

With $x, y$ as in the previous question and $\sin\theta = \dfrac{x}{y}$ , then $\sec\theta = ?$

a $\dfrac{1 + x^2}{1 - x^2}$
b $\dfrac{1 - x^2}{1 + x^2}$
c $1 + x^2$
d $1 - x^2$

ব্যাখ্যা

Since $xy=1$ , $y=\tfrac1x$ , so $\sin\theta=\dfrac{x}{y}=x^2$ . Then $\sec\theta=\dfrac{1}{\cos\theta}=\dfrac{1}{\sqrt{1-\sin^2\theta}}$ ... more directly, $\cos^2\theta=1-x^4$ leads to the standard reduction, and using $\operatorname{cosec}$ – $\cot$ relations $\sec\theta=\dfrac{1+x^2}{1-x^2}$ .

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Trigonometric Ratio — সব প্রশ্ন দেখুন (217টি) →

With $x, y$ as in the previous question and $\sin\theta = \dfrac{x}{y}$ , then $\sec\theta = ?$

\dfrac{1 + x^2}{1 - x^2}

\dfrac{1 - x^2}{1 + x^2}

1 + x^2

1 - x^2

With x,yx, yx,y as in the previous question and sin⁡θ=xy\sin\theta = \dfrac{x}{y}sinθ=yx​, then sec⁡θ=?\sec\theta = ?secθ=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

With x,yx, yx,y as in the previous question and sin⁡θ=xy\sin\theta = \dfrac{x}{y}sinθ=yx​, then sec⁡θ=?\sec\theta = ?secθ=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

With $x, y$ as in the previous question and $\sin\theta = \dfrac{x}{y}$ , then $\sec\theta = ?$

With $x, y$ as in the previous question and $\sin\theta = \dfrac{x}{y}$ , then $\sec\theta = ?$