Question

under root 1 + sin theta ÷1-sin theta =sec theta +tan theta​

Answer 1

QUESTION:

$\sf Prove\ \ \sqrt{ \dfrac{1 + sin \ \theta }{1-sin \ \theta} } = sec \ \theta +tan \ \theta$

TO PROVE:

$\sf\sqrt{ \dfrac{1 + sin \ \theta }{1-sin \ \theta} } = sec \ \theta +tan \ \theta$

PROOF:

$\sf Take\ {\bf{\sqrt{ \dfrac{1 + sin \ \theta }{1-sin \ \theta} } }}\ \ as \ L.H.S$

$\sf Take\ { \bf{ sec \ \theta +tan \ \theta \ }} \ as \ R.H.S$

L.H.S:

$\tt{ \sqrt{ \dfrac{1 + sin \ \theta }{1-sin \ \theta} } }$

$\tt{Multiply\ and\ divide\ by\ \bf\sqrt{1 + sin\ \theta}}$

$\tt{ \sqrt{ \dfrac{1 + sin \ \theta }{1-sin \ \theta} \times \dfrac{1 + sin\ \theta}{1 + sin\ \theta}} }$

$\blue{\boxed{\bold{\bigstar\gray{( A+ B)(A-B) = A ^{2} - B^{2} }}\bigstar}}$

$\tt{ \sqrt{ \dfrac{(1 + sin \ \theta )^{2} }{1^{2} -sin ^{2} \ \theta} }}$

$\red{\boxed{\bold{\bigstar\gray{ 1 - {sin}^{2} \theta = {cos}^{2} \theta }}\bigstar}}$

$\tt{ \sqrt{ \dfrac{(1 + sin \ \theta )^{2} }{cos ^{2} \ \theta} }}$

$\tt{ \dfrac{1 + sin \ \theta }{cos \ \theta} }$

$\large \tt{\underline{\underline{Split\ the\ numerator.}}}$

$\tt{ \dfrac{1}{ cos \ \theta } } + \left( \tt{\dfrac{ sin \ \theta }{cos \ \theta} } \right)$

$\pink{\boxed{\bold{\bigstar\gray{ \dfrac{1}{ cos \ \theta } = sec \ \theta} \bigstar }}}$

$\orange{\boxed{\bold{\bigstar\gray{\left( {\dfrac{ sin \ \theta }{cos \ \theta} } \right) = tan \ \theta} \bigstar}}}$

$\tt{ \dfrac{1}{ cos \ \theta } } \left( \tt{\dfrac{ sin \ \theta }{cos \ \theta} } \right) =sec \ \theta + tan \ \theta$

L.H.S:

$\tt { sec \ \theta + tan \ \theta \ }$

L.H.S = R.H.S

$\sf\sqrt{ \dfrac{1 + sin \ \theta }{1-sin \ \theta} } = sec \ \theta + tan \ \theta$

HENCE PROVED.

Answer 2

$\huge\pink{\underline{\boxed{\mathbb{Questions?}}}}$]

under root 1 + sin theta ÷1-sin theta =sec theta +tan theta❓❓

$\huge\pink{\underline{\boxed{\mathbb{AnSWeR}}}}$]

✎Firstly consider L.H.S ; NOW rationalise

✎ √ 1 +SinA √ 1+ SinA

--------------- x ----------------

√ 1- SinA √1 + SinA

✎√ ( 1+ SinA)² 1+ SinA

------------------ = --------------

√ 1² - Sin²A Cos A

✎ Now consider RHS

✎= Sec A + Tan A

✎=1/ Cos A + SinA / CosA

✎=1 + SinA / CosA

$\huge\green{\underline{\boxed{\mathbb{Hence ,proved}}}}$]