Question

plz answer the question. ​

Answer 1

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

$\red{\underline \bold{To \: Prove:}} \\ \tt: \implies \frac{1 - {tan}^{2} \: \theta}{ {cot}^{2} \: \theta - 1 } = {tan}^{2} \: \theta$

• According to given question :

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

Answer 2

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

QUESTION :

$Prove \: That \: :- \dfrac{1 - {tan { \phi }}^2 } { { cot{ \phi} ^2 } - 1 } = {tan {\phi } }^2$

SOLUTION :

LHS :

$\tt{\purple{\leadsto{ \dfrac{1 - {tan { \phi }}^2 } { { cot{ \phi} ^2 } - 1 } }}}$

$\tt{\orange{\leadsto{ \dfrac{1 - { \dfrac{ sin { \phi } } { cos { \phi }} }^2 } { { \dfrac{ cos { \phi }} {sin { \phi}}}^2 - 1 } }}}$

Take LCM and cancel the Numerator Sin ^2 Phi - Cos ^ Phi

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

$\tt{\red{\leadsto { {tan {\phi } }^2 }}}$

LHS = RHS

HENCE PROVED.....