Question

If tan θ = $\frac{a}{x}$, find the value of $\frac{x}{ \sqrt{ {a}^{2} + {x}^{2} } }$​

Answer 1

Given that,

$\rm\tan\theta=\dfrac{a}{x}$

$\;$

Given expression in terms of trigonometric ratio is,

$\rm\dfrac{x}{\sqrt{a^{2}+x^{2}}}$

$\;$

$\rm=\dfrac{x}{\sqrt{a^{2}+x^{2}}}\times\dfrac{\dfrac{1}{a}}{\dfrac{1}{a}}$

$\;$

$\rm=\dfrac{\dfrac{x}{a}}{\sqrt{\dfrac{1}{a^{2}}(a^{2}+x^{2})}}$

$\;$

$\rm=\dfrac{\dfrac{x}{a}}{\sqrt{1+\dfrac{x^{2}}{a^{2}}}}$

$\;$

$\rm=\dfrac{\dfrac{1}{\tan\theta}}{\sqrt{1+\dfrac{1}{\tan^{2}\theta}}}$

$\;$

$\rm=\dfrac{\dfrac{1}{\tan\theta}}{\sqrt{\dfrac{\tan^{2}\theta+1}{\tan^{2}\theta}}}$

$\;$

$\boxed{\rm\tan^{2}+1=\sec^{2}\theta}$

$\rm=\dfrac{\dfrac{1}{\tan\theta}}{\sqrt{\dfrac{\sec^{2}\theta}{\tan^{2}\theta}}}$

$\;$

$\rm=\dfrac{\dfrac{1}{\tan\theta}}{\sqrt{\left(\dfrac{\sec\theta}{\tan\theta}\right)^{2}}}$

$\;$$\boxed{\rm\sqrt{a^{2}}=a\ (a\geq0)}$

$\rm=\dfrac{\dfrac{1}{\tan\theta}}{\dfrac{\sec\theta}{\tan\theta}}$

$\;$

$\rm=\dfrac{1}{\sec\theta}$

$\;$

$\rm=\cos\theta$

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$\rm\therefore\dfrac{x}{\sqrt{a^{2}+x^{2}}}=\cos\theta$

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$\Large\textrm{Learn More}$

$\textbf{- Trigonometic Identities}$

$\boxed{\rm{\sin^{2}\theta+\cos^{2}\theta=1}}$

$\;$

$\boxed{\rm{\tan^{2}\theta+1=\sec^{2}\theta}}$

$\;$

$\boxed{\rm \tan\theta=\dfrac{\sin\theta}{\cos\theta}}$

$\;$

$\textbf{- Reciprocal Ratios}$

$\boxed{\rm{\sin\theta\csc\theta=1}}$

$\;$

$\boxed{\rm{\cos\theta\sec\theta=1}}$

$\;$

$\boxed{\rm{\tan\theta\cot\theta=1}}$