Question

root under sec squared theta + cos squared theta is equal to sec theta into cosec theta prove that​

Answer 1

Solution →

$\sqrt{ {sec}^{2} \: \theta + cos</strong><strong>e</strong><strong>c</strong><strong>^{2} \: \theta } = \sec \:\theta \times cosec \: \: \theta$

LHS →

$= \sqrt{ {sec}^{2} \: \theta + cos</strong><strong>e</strong><strong>c</strong><strong>^{2} \: \theta }$

we know that →

• $sec^{2}\theta = \frac{1}{ {cos}^{2}\theta }$
• $cos</strong><strong>e</strong><strong>c</strong><strong>^{2}\theta = \frac{1}{ {</strong><strong>sin</strong><strong>}^{2}\theta }$

Putting tha value of

${</strong><strong>cosec</strong><strong>}^{2} \theta$ and ${sec^{2}\theta}$

→

$\sqrt{ \frac{1}{ {cos}^{2}\theta } + \frac{1}{ {sin}^{2}\theta } }$

$= \sqrt{ \frac{ {sin}^{2}\theta + {cos}^{2}\theta }{ {cos}^{2}\theta \times {sin}^{2}\theta } }$

We know that →

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

Then,

$= \sqrt{ \frac{1}{ {cos}^{2}\theta \times {sin}^{2}\theta } }$

$= \frac{1}{sin \: \theta \times cos\theta}$

We can write it as ,

$= cosec \: \theta \times \sec \: \theta$

$\boxed{=sec \: \theta × cosec\: \theta}$ RHS

Hence Proved !!

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