Question

यदि tan A=y/3 तो sec A का मान क्या होगा​

Answer 1

Answer:

$SecA=\frac{\sqrt{9+y^{2}}}{3}$

Step-by-step explanation:

$Given \: tanA = \frac{y}{3}$

$SecA = \sqrt{1+tan^{2}A}$

$=\sqrt{1+\left(\frac{y}{3}\right)^{2}}$

$=\sqrt{1+\frac{y^{2}}{9}}$

$=\sqrt{\frac{9+y^{2}}{9}}$

$=\frac{\sqrt{9+y^{2}}}{3}$

Therefore,

$SecA=\frac{\sqrt{9+y^{2}}}{3}$

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Answer 2

Answer:

Step-by-step explanation: