Question

To Prove :
$\small \tt2 {sec}^{2} θ - {sec}^{4} θ - 2 {cosec}^{2} θ + {cosec}^{4} θ = {cot}^{4} - {tan}^{4}$


Only For :

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Answer 1

GiveN :

$\tt \dagger \: \: \: \: \: 2sec^2\theta - Sec^4 \theta - 2 Cosec^2\theta + Cosec^4\theta = Cot^4\theta - tan^4\theta$

To ProvE :

LHS = RHS.

SolutioN :

$\tt : \implies 2sec^2\theta - Sec^4 \theta - 2 Cosec^2\theta + Cosec^4\theta$

$\tt : \implies-(-2sec^2\theta + Sec^4 \theta) - 2 Cosec^2\theta + Cosec^4\theta$

$\tt : \implies Cosec^4\theta- 2 Cosec^2\theta -(-2sec^2\theta + Sec^4 \theta)$

$\tt : \implies ( Cosec^4\theta- 2 Cosec^2\theta + 1 ) - (Sec^4\theta -2sec^2\theta+1)$

$\tt : \implies ( Cosec^2\theta- 1 )^2 - (Sec^2\theta -1)^2$

$\tt : \implies (Cot^2\theta)^2 - (tan^2\theta)^2$

$\tt : \implies Cot^4\theta - tan^4\theta$

Hence Proved.

_______________________________

Formula Use :

• ( a ± b ) = a² ± 2ab + b².

• 1 + tan²A = Sec²A.

• 1 + Cot²A = Cosec²A.

Answer 2

Answer:

$\sf\boxed{\bold{\large{Required\: Solution :-}}}$

$\sf \\ \longmapsto\sf\bold{\red{L.H.S}}$

$\sf \\ \implies 2 {sec}^{2}\theta -\: {sec}^{4}\theta - 2 {cosec}^{2}\theta + {cosec}^{4}\theta$

$\sf \\ \implies 2 {sec}^{2}\theta -\: ({sec}^{2}\theta)^{2} -\: 2 {cosec}^{2}\theta + ({cosec}^{2}\theta)^{2}$

As we know that,

$\leadsto$ $\sf 1 +\: {tan}^{2}\theta =\: {sec}^{2}\theta$

$\leadsto$ $\sf 1 +\: {cot}^{2}\theta =\: {cosec}^{2}\theta$

$\sf \\ \implies 2(1 + {tan}^{2}\theta) - (1 + {tan}^{2}\theta)^{2} - 2(1 + {cot}^{2}\theta) + (1 + {cot}^{2}\theta)^{2}$

$\sf \\ \implies 2 + 2{tan}^{2}\theta - (1 + 2{tan}^{2}\theta + {tan}^{4}\theta) - 2 - 2{cot}^{2}\theta + 1 + 2{cot}^{2}\theta + {cot}^{4}\theta$

$\sf \\ \implies {\cancel{2}} + {\cancel{2{tan}^{2}\theta}} {\cancel{- 1}} {\cancel{- 2{tan}^{2}\theta}} - {tan}^{4}\theta {\cancel{- 2}} {\cancel{- 2{cot}^{2}\theta}} {\cancel{+ 1}} {\cancel{+ 2{cot}^{2}\theta}} + {cot}^{4}\theta$

$\sf \\ \implies - {tan}^{4}\theta + {cot}^{4}\theta$

$\sf \\ \implies \bold{\purple{{cot}^{4}\theta - {tan}^{4}\theta}}$

$\sf \\ \longmapsto \bold{\red{R.H.S}}$

$\sf \\ \implies \bold{\pink{L.H.S = R.H.S}}$

${\underline{\boxed{\large{\bf{HENCE,\: PROVED}}}}}$