Question

prove that (sin^4teta-cos^4teta+1)cosec^2teta=2

Answer 1

Answer:

(sin⁴θ-cos⁴θ+1)cosec²θ

=[{(sin²θ)²-(cos²θ)²}+1]cosec²θ

=[{(sin²θ+cos²θ)(sin²θ-cos²θ)}+1]cosec²θ

=(sin²θ-cos²θ+1)cosec²θ [∵, sin²θ+cos²θ=1]

={sin²θ+(1-cos²θ)}cosec²θ

=(sin²θ+sin²θ)cosec²θ

=2sin²θ.cosec²θ

=2sin²θ×1/sin²θ

=2 (Proved)

pls mark me brainliest

Step-by-step explanation:

Answer 2

${ \bold{ \underline{ TO \: \: PROVE } : - }} \\ \\ { \bold{ ({sin}^{4}(\theta) - {cos}^{4}{ (\theta) + 1) {cosec}^{2}{ \theta} } = 2 }}$

${ \bold{ \underline{ SOLUTION } : - }} \\ \\ \\ { \bold{ \underline{ L.H.S.} : } - } \\ \\ { \bold{ = ({sin}^{4}{ \theta} - {cos}^{4}{ \theta} + 1) {cosec}^{2} { \theta} }} \\ \\ { \bold{ = ({sin}^{4}{ \theta} - {cos}^{4}{ \theta} + 1) \times ( \frac{1}{ {sin}^{2} { \theta}}) }} \\ \\ { \bold{ \: \: \: \: . \: \: we \: \: know \: \: that \: - }} \\ \\ { \bold{ \implies \: {a}^{2} - {b}^{2} = (a + b)(a - b) }} \\ \\ { \bold{ = ( ( {sin}^{2}{ \theta} + {cos}^{2} { \theta})( {sin}^{2}{ \theta} - {cos}^{2} { \theta}) + 1) \times \frac{1}{ { \sin }^{2} { \theta}} }} \\ \\ { \bold{ =( {sin}^{2}{ \theta} - {cos}^{2} { \theta} + 1) \times \frac{1}{ { \sin }^{2} { \theta}} }} \\ \\ { \bold{ = ( {sin}^{2} { \theta} + {sin}^{2}{ \theta}) \times \frac{1}{ {sin}^{2}{ \theta} } }} \\ \\ { \bold{ = 2 \times \cancel \frac{{sin}^{2} { \theta}}{ {sin}^{2}{ \theta} } }} \\ \\ { \bold{ = 2}} \\ \\ { \bold{ = R.H.S. \: \: \: \: ( Hence \: \: proved) }}$