Question

Prove that: (sin4θ-cos4θ+1)cosec2θ=2

Answer 1

HEY Buddy......!! here is ur answer

we have to prove that :

$( { \sin}^{4} \alpha - { \cos }^{4} \alpha + 1) { \csc }^{2} \alpha = 2 \\ \\ On \: taking \: L.H.S. \\ \\ ( { \sin}^{4} \alpha - { \cos }^{4} \alpha + 1) { \csc }^{2} \alpha \\ \\ = > [( { \sin}^{2} \alpha - { \cos }^{2} \alpha )( { \sin }^{2} \alpha + { \cos }^{2} \alpha ) + 1] { \csc }^{2} \alpha \\ \\ As \: we \: know \: that... \: {a}^{2} - {b}^{2} = (a + b)(a - b) \: and \: { \sin }^{2} \alpha + { \cos}^{2} \alpha = 1 \\ \\ = > ( { \sin }^{2} \alpha - { \cos }^{2} \alpha + 1) { \csc }^{2} \alpha \\ \\ = > 2 { \sin }^{2} \alpha { \csc}^{2} \alpha \\ \\ = > 2 = \: R.H.S. \\ \\ As \: we \: know \: that....sin \alpha . \csc\alpha = 1$

I hope it will be helpful for you....!!

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Answer 2

To prove that:  ($\sin^4 \theta$ - $\cos^4 \theta$ + 1)$\csc^2 \theta$ = 2.

Solution:

L.H.S. = ($\sin^4 \theta$ - $\cos^4 \theta$ + 1)$\csc^2 \theta$

= ($(\sin^2 \theta)^2$ - $(\cos^2 \theta)^2$ + 1)$\csc^2 \theta$

Using the algebraic identity,

$a^{2} -b^{2}$ = (a + b)(a - b)

= (($\sin^2 \theta$ + $\cos^2 \theta$)($\sin^2 \theta$ - $\cos^2 \theta$)+ 1)$\csc^2 \theta$

Using the trigonometric identity,

$\sin^2 A$ + $\cos^2 A$ = 1

= ((1)($\sin^2 \theta$ - $\cos^2 \theta$)+ 1)$\csc^2 \theta$

= (($\sin^2 \theta$ - $\cos^2 \theta$)+ 1)$\csc^2 \theta$

Using the trigonometric identity,

$\cos^2 A$ = 1 - $\sin^2 A$

= (($\sin^2 \theta$ - (1 - $\sin^2 \theta$ ) + 1)$\csc^2 \theta$

= ($\sin^2 \theta$ - 1 + $\sin^2 \theta$  + 1)$\csc^2 \theta$

= (2 $\sin^2 \theta$ )$\csc^2 \theta$

Using the trigonometric identity,

$\csc A =\dfrac{1}{\sin A}$

= (2 $\sin^2 \theta$ )$\dfrac{1}{\sin^2 \theta}$

= 2

= R.H.S., proved.

Thus, ($\sin^4 \theta$ - $\cos^4 \theta$ + 1)$\csc^2 \theta$ = 2, proved.