Question

Tan square alpha + cot square alpha + 2 = sec square alpha.cosec square alpha​

Answer 1

To Prove :-

$\sf {tan}^{2} \alpha + {cot}^{2} \alpha + 2 = {sec}^{2} \alpha {cosec}^{2} \alpha$

Solution :-

$: \implies \sf LHS = {tan}^{2} \alpha + {cot}^{2} \alpha + 2$

$\bf \dag \: tan \alpha = \dfrac{sin \alpha}{cos \alpha}$

$\bf \dag \: cot \alpha = \dfrac{cos \alpha}{sin \alpha}$

$: \implies \sf LHS = \dfrac{ {sin}^{2} \alpha}{ {cos}^{2} \alpha } + \dfrac{ {cos}^{2} \alpha }{ {sin}^{2} \alpha } + 2$

$: \implies \sf LHS = \dfrac{ {sin}^{4} \alpha + {cos}^{4} \alpha }{ {sin}^{2} \alpha {cos}^{2} \alpha} + 2$

$: \implies \sf LHS = \dfrac{ {sin}^{4} \alpha + {cos}^{4} \alpha + 2 {sin}^{2} \alpha {cos}^{2} \alpha}{ {sin}^{2} \alpha {cos}^{2} \alpha }$

$: \implies \sf LHS = \dfrac{( {sin}^{2} \alpha + {cos}^{2} \alpha ) ^{2} }{ {sin}^{2} \alpha {cos}^{2} \alpha}$

$\bf \dag \: {sin}^{2} \alpha + {cos}^{2} \alpha = 1$

$: \implies \sf LHS = \dfrac{1}{ {sin}^{2} \alpha {cos}^{2} \alpha}$

$\bf \dag \: sec \alpha = \dfrac{1}{cos \alpha}$

$\bf \dag \: cosec \alpha = \dfrac{1}{sin \alpha}$

$: \implies \sf LHS = {sec}^{2} \alpha {cosec}^{2} \alpha$

$: \implies \sf LHS = RHS$

Hence proved.

Answer 2

${ \boxed{\underline{\tt{ \orange{Required \: answer:-}}}}}$

★TO PROVE:-

$: \implies \sf \bf{Tan^2 \alpha + Cot^2 \alpha +2 = \: Sec^2 \alpha × Cosec^2 \alpha \: }$

★FORMULA USED:-

$\bigcirc \: \: \: \: \sf \bf{Tan\theta = \frac{ Sin \theta }{Cos \theta} }$

$\bigcirc \: \: \: \sf \bf{Cot \theta = \frac{Cos\theta}{Sin\theta} }$

$\bigcirc \: \: \: \: \sf \bf{ \ { \sin }^{2} \theta \: + { \cos }^{2} \theta = 1 }$

$\bigcirc \: \: \: \: \sf \bf{ \frac{1}{sin \theta} = cosec \: \theta \: }$

$\bigcirc \: \: \: \: \sf \bf{} \frac{1}{cos \theta } = sec \theta \:$

★SOLUTION:-

${\boxed{\underline{\bf{\red{ L.H.S}}}}}$

$: \implies \sf{Tan^2 \alpha + Cot^2 \alpha +2}$

$:\implies\sf{ \frac{{sin}^{2} \theta}{ {cos}^{2} \theta } + \frac{ {cos}^{2} \theta}{ {sin}^{2} \theta} + 2 }$

$: \implies \sf{ \frac{ {sin}^{4} \theta \: + {cos}^{4} \theta \: + 2 \: {cos}^{2} \theta \: {sin}^{2} \theta }{ {cos}^{2} \theta \: {sin}^{2} \theta \: } }$

$: \implies \sf{ \frac{ {( {sin}^{2} \theta \: + {cos}^{2} \theta)}^{2} }{ {sin}^{2} \theta \: {cos}^{2} \theta} }$

$: \implies \sf{ \frac{1}{ {sin}^{2} \theta \: {cos}^{2} \theta \: } }$

$: \longrightarrow \sf{ {cosec}^{2} \theta \: . \: {sec}^{2} \theta \: }$

${\boxed{\underline{\bf{\red{R .H.S}}}}}$

So, LHS = RHS

HENCE PROVED.