Question

plz solve this question and answer it here . don't try to take the points just by telling hello or don't tell that the question is proper or not . I will report u guys ​

Answer 1

Given Question :- Prove that

$\rm \: {2sec}^{2}\theta - {sec}^{4}\theta - {2cosec}^{2}\theta + {cosec}^{4}\theta = {cot}^{4}\theta - {tan}^{4}\theta$

$\large\underline{\sf{Solution-}}$

Consider RHS

$\rm \: {cosec}^{4}\theta - {cot}^{4}\theta$

$\rm \: = \: {( {cot}^{2} \theta )}^{2} - {( {tan}^{2} \theta )}^{2}$

We know,

$\boxed{ \rm{ \: {cosec}^{2}x - {cot}^{2}x = 1 \: }} \\ \\ \rm \: and \\ \\ \boxed{ \rm{ \: {sec}^{2}x - {tan}^{2}x = 1 \: }}$

So, using these identities, we get

$\rm \: = \: {( {cosec}^{2}\theta - 1)}^{2} - {( {sec}^{2} \theta - 1)}^{2}$

We know,

$\boxed{ \rm{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \: }}$

So, using this identity, we get

$\rm \: = \: {cosec}^{4}\theta + 1 - {2cosec}^{2}\theta - ( {sec}^{4}\theta + 1 - 2 {sec}^{2}\theta )$

$\rm \: = \: {cosec}^{4}\theta + 1 - {2cosec}^{2}\theta - {sec}^{4}\theta - 1 + 2 {sec}^{2}\theta$

$\rm \: = \: {cosec}^{4}\theta - {2cosec}^{2}\theta - {sec}^{4}\theta+ 2 {sec}^{2}\theta$

can be re-arranged as

$\rm \: = \: {2sec}^{2}\theta - {sec}^{4}\theta - {2cosec}^{2}\theta + {cosec}^{4}\theta$

Hence,

$\boxed{ \rm{ \:{2sec}^{2}\theta- {sec}^{4}\theta - {2cosec}^{2}\theta+{cosec}^{4}\theta= {cot}^{4}\theta- {tan}^{4}\theta \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 }\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Answer:

Given Question :- Prove that

$\rm \: {2sec}^{2}\theta - {sec}^{4}\theta - {2cosec}^{2}\theta + {cosec}^{4}\theta = {cot}^{4}\theta - {tan}^{4}\theta$

$\large\underline{\sf{Solution-}}$

Consider RHS

$\rm \: {cosec}^{4}\theta - {cot}^{4}\theta$

$\rm \: = \: {( {cot}^{2} \theta )}^{2} - {( {tan}^{2} \theta )}^{2}$

We know,

$\boxed{ \rm{ \: {cosec}^{2}x - {cot}^{2}x = 1 \: }} \\ \\ \rm \: and \\ \\ \boxed{ \rm{ \: {sec}^{2}x - {tan}^{2}x = 1 \: }}$

So, using these identities, we get

$\rm \: = \: {( {cosec}^{2}\theta - 1)}^{2} - {( {sec}^{2} \theta - 1)}^{2}$

We know,

$\boxed{ \rm{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \: }}$

So, using this identity, we get

$\rm \: = \: {cosec}^{4}\theta + 1 - {2cosec}^{2}\theta - ( {sec}^{4}\theta + 1 - 2 {sec}^{2}\theta )$

$\rm \: = \: {cosec}^{4}\theta + 1 - {2cosec}^{2}\theta - {sec}^{4}\theta - 1 + 2 {sec}^{2}\theta$

$\rm \: = \: {cosec}^{4}\theta - {2cosec}^{2}\theta - {sec}^{4}\theta+ 2 {sec}^{2}\theta$

can be re-arranged as

$\rm \: = \: {2sec}^{2}\theta - {sec}^{4}\theta - {2cosec}^{2}\theta + {cosec}^{4}\theta$

Hence,

$\boxed{ \rm{ \:{2sec}^{2}\theta- {sec}^{4}\theta - {2cosec}^{2}\theta+{cosec}^{4}\theta= {cot}^{4}\theta- {tan}^{4}\theta \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 }\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$