Question

Prove that
(sin4theta - cos4theta + 1)cosec²theta = 2 ​

Answer 1

GIVEN :

• (sin4∅ - cos4∅ + 1) cosec²∅ = 2.

TO PROVE :-

• L.H.S = R.H.S.

SOLUTION :-

L.H.S,

$\\ : \implies \displaystyle\sf \: \bigg( \sin4 \theta - \cos \theta + 1 \bigg) \csc^{2} \theta$

$: \implies \displaystyle\sf \: \Bigg[ \bigg( \sin ^{2} \theta \bigg) ^{2} - \bigg( \cos^{2} \theta \bigg) ^{2} + 1 \Bigg ]\csc^{2} \theta$

$\bullet \displaystyle\sf \:using \: identity \: \div \: a ^{2} - b ^{2} = (a + b)(a - b)$

$: \implies \displaystyle\sf \:\Bigg[ \bigg( \sin ^{2} \theta + \cos^{2} \theta\bigg ) \bigg(\sin ^{2} \theta - \cos^{2} \theta \bigg) + 1 \Bigg ]\csc^{2} \theta$

$\bullet \: \displaystyle\sf \: using \: identity \: \div \: \sin ^{2} \theta + \cos^{2} \theta = 1$

$: \implies \displaystyle\sf \Bigg[ 1 \bigg(\sin ^{2} \theta - \cos^{2} \theta \bigg) + 1 \Bigg ]\csc^{2} \theta$

$: \implies \displaystyle\sf \Bigg[ \bigg(\sin ^{2} \theta - \cos^{2} \theta \bigg) + 1 \Bigg ]\csc^{2} \theta$

$: \implies \displaystyle\sf \Bigg[ \sin ^{2} \theta + \bigg(1- \cos^{2} \theta \bigg) \Bigg ]\csc^{2} \theta$

$\bullet \: \displaystyle\sf using \: identity \: \div \: \sin ^{2} \theta = 1- \cos^{2} \theta$

$: \implies \displaystyle\sf \Bigg[ \sin ^{2} \theta + \sin ^{2} \theta \Bigg ]\csc^{2} \theta$

$: \implies \displaystyle\sf \Bigg[ \bigg(2 \sin ^{2} \theta \bigg) \bigg( \csc ^{2} \theta \bigg) \Bigg ]$

$\bullet \: \displaystyle\sf using \: identity \: \csc ^{2} \theta = \dfrac{1}{ \sin ^{2} \theta}$

$: \implies \displaystyle\sf \: 2 \cancel{\sin \theta \times } \dfrac{1}{ \cancel{\sin ^{2} \theta}}$

$: \implies \underline{ \boxed{ \displaystyle\sf2}} \: \: \: \: \: \: \: \: \: \: \: \bigg \lgroup \displaystyle\sf \: L.H.S\bigg \rgroup$

________________________

$\\ \\ \dashrightarrow \bigg \lgroup \displaystyle\sf \: L.H.S\bigg \rgroup = \bigg \lgroup \displaystyle\sf \: R.H.S\bigg \rgroup$

HENCE PROVED.

Answer 2

Answer:

Step-by-step explanation:

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(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)