Question

can anyone solve this step by step​

Answer 1

$\large \mathbb \blue{TO \: PROVE : - }$

$\sf \: \frac{1}{sec \theta + tan \theta} = {sec \theta + tan \theta}$

$\large \mathbb \blue{SOLUTION: - }$

$\sf \: \: L.H.S. \: \: = \frac{1}{sec \theta + tan \theta}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: = \frac{{sec \theta + tan \theta}}{({sec \theta - tan \theta}) ({sec \theta + tan \theta}) }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: = \frac{sec \theta + tan \theta}{ {sec}^{2} \theta - {tan}^{2} \theta }$

$\star \: \rm \red{\: by \: algebric \: identity}$

$\green{ \boxed{ \sf \: (a + b)(a - b) = {a}^{2} - {b}^{2} }}$

$\longrightarrow \sf \: \frac{sec \theta + tan \theta}{1}$

$\star \: \rm \red{\: by \: trignometric \: identity}$

$\sf \green{\boxed { \sf {sec}^{2} A - {tan}^{2} A = 1}}$

$\implies \sf \: sec \theta + tan \theta = R.H.S.$

$\tt \: therefore.$

$\boxed{\sf \: \frac{1}{sec \theta + tan \theta} = {sec \theta + tan \theta} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frak \purple{ \: hence \: proved} \: \: \: \: \: \: \:$

Answer 2

please mark it as brainliest answer