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If tan theeta =a/x then x/root off a square +xsqure

Posted by Noufal Kasala 2 years, 2 months ago

CBSE > Class 10 > Mathematics

1 answers

Preeti Dabral 2 years, 2 months ago

Value of $\frac{x}{\sqrt{a^2+x^2}}$

$\tan \theta=\frac{\mathrm{a}}{\mathrm{x}} Since, \begin{aligned} & \frac{x}{\sqrt{a^2+x^2}} \\ & \Rightarrow \frac{\frac{a}{\tan \theta}}{\sqrt{a^2+\left(\frac{a}{\tan \theta}\right)^2}} \\ & \Rightarrow \frac{\frac{a}{\tan \theta}}{\sqrt{a^2+\frac{a^2}{\tan ^2 \theta}}} \end{aligned}$

$\begin{aligned} & \Rightarrow \frac{\frac{a}{\tan \theta}}{a \sqrt{\frac{1+\tan ^2 \theta}{\tan ^2 \theta}}} \\ & \Rightarrow \frac{\frac{1}{\tan \theta}}{\sqrt{\frac{1+\tan ^2 \theta}{\tan ^2 \theta}}} \\ & \Rightarrow \frac{1}{\sqrt{\sec ^2 \theta}} \\ & \Rightarrow \frac{1}{\sec \theta} \end{aligned}$

$\Rightarrow \cos \theta$

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