Question

plz answer the question. ​

Answer 1

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\red{\underline \bold{To \: Prove:}} \\ \tt: \implies \frac{1 -cos \: \theta}{1 + cos \: \theta} = ( {cosec \: \theta -cot \: \theta)}^{2}$

• According to given question :

$\tt: \implies \frac{1 -cos \: \theta}{1+ cos \: \theta} = ( {cosec \: \theta -cot \: \theta)}^{2} \\ \\ \bold{Solving \: L.H.S} \\ \tt: \implies \frac{1 +cos \: \theta}{1- cos \: \theta} \\ \\ \tt \circ \: Multiplying \: \: \frac{1 - cos \: \theta}{1 - cos \: \theta} \\ \\ \tt: \implies \frac{1 - cos \: \theta}{1 +cos \: \theta} \times \frac{1 -cos \: \theta}{1 - cos \: \theta} \\ \\ \tt: \implies \frac{(1 - cos \: \theta)^{2} }{ {1}^{2} - cos^{2} \: \theta} \\ \\ \tt: \implies \frac{(1 - cos \: \theta)^{2} }{ {sin}^{2} \: \theta } \\ \\ \tt: \implies (\frac{1 -cos \: \theta}{sin \: \theta})^{2} \\ \\ \tt: \implies (\frac{1}{sin \: \theta} - \frac{cos \: \theta}{sin \: \theta} )^{2} \\ \\ \green{\tt: \implies (cosec \: \theta -cot \: \theta)^{2}} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \green{L.H.S =R.H.S} \\ \\ \: \: \: \: \: \: \: \: \red{\huge{ \bold{Proved}}}$

Answer 2

$\tt{\huge{\orange {Hello!!! }}} B.Q.$

$\tt{\purple{\leadsto{ \dfrac{ 1 + cos { \phi } } { 1 - cos { \phi } } }}}$

Rationalizing,

$\tt{\orange{\leadsto{ \dfrac{ 1 + cos { \phi } } { 1 - cos { \phi } } \times \dfrac{ 1 + cos { \phi } } { 1 + cos { \phi } } }}}$

$\tt{\blue{\leadsto{ \dfrac { { 1 + cos { \phi } }^2 }{1 - { cos { \phi }}}^2 } }}}$

$\tt{\green{\leadsto{ \dfrac { { 1 + cos { \phi } }^2 }{ {sin { \phi }}^2 } }}}$

$\tt{\red{\leadsto{ \dfrac{ 1 + cos {\phi} { sin {\phi} }^2 }}}$

$\tt{\orange{\leadsto{ { Cosec { \phi } + Cot { \phi} } ^2 }}}$

HENCE PROVED