Question

root under sec square A + cosec square A = secA . cosec A​

Answer 1

under root sec square A =secA

under root cosec square A=CosecA

=SecA.CosecA=RHS

Hope it helps

Mark it as BRAINLIEST if you are satisfied with the answer

Answer 2

Solution →

$\sqrt{ {sec}^{2} \: \theta + cosec^{2} \: \theta } = \sec \:\theta \times cosec \: \: \theta$

LHS →

$= \sqrt{ {sec}^{2} \: \theta + cosec^{2} \: \theta }$

we know that →

• $sec^{2}\theta = \frac{1}{ {cos}^{2}\theta }$
• $cosec^{2}\theta = \frac{1}{ {sin}^{2}\theta }$

Putting tha value of

${cosec}^{2} \theta$ 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 !!

__________________________

───── ❝ TheEnforceR ❞ ─────

Mark as Brainliest !! ✨✨