Question

If (cos⁴α sec²β) + (sin⁴α cosec²β) = 1, prove that sin⁴α + sin⁴β = 2 sin²α sin²β

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Answer 1

ANSWER:

Given:

• (cos⁴α sec²β) + (sin⁴α cosec²β) = 1

To Prove:

• sin⁴α + sin⁴β = 2 sin²α sin²β

Solution:

We are given that,

$\implies(\cos^4\alpha\sec^2\beta)+(\sin^4\alpha\csc^2\beta)=1$

We know that,

$\hookrightarrow\sec\theta=\dfrac{1}{\cos\theta}$

And,

$\hookrightarrow\csc\theta=\dfrac{1}{\sin\theta}$

So,

$\implies(\cos^4\alpha\sec^2\beta)+(\sin^4\alpha\csc^2\beta)=1$

$\implies\dfrac{\cos^4\alpha}{\cos^2\beta}+\dfrac{\sin^4\alpha}{\sin^2\beta}=1$

$\implies\dfrac{(\cos^2\alpha)^2}{\cos^2\beta}+\dfrac{\sin^4\alpha}{\sin^2\beta}=1$

We know that,

$\hookrightarrow \cos^2\theta=1-\sin^2\theta$

So,

$\implies\dfrac{(\cos^2\alpha)^2}{\cos^2\beta}+\dfrac{\sin^4\alpha}{\sin^2\beta}=1$

$\implies\dfrac{(1-\sin^2\alpha)^2}{(1-\sin^2\beta)}+\dfrac{\sin^4\alpha}{\sin^2\beta}=1$

Taking LCM,

$\implies\dfrac{[(\sin^2\beta)\times(1-\sin^2\alpha)^2]+[(1-\sin^2\beta)\times(\sin^4\alpha)]}{(1-\sin^2\beta)(\sin^2\beta)}=1$

$\implies\dfrac{[(\sin^2\beta)\times(1^2+\sin^4\alpha-2\sin^2\alpha)]+[(1-\sin^2\beta)\times(\sin^4\alpha)]}{(1-\sin^2\beta)(\sin^2\beta)}=1$

Transposing denominator to RHS,

$\implies\dfrac{[(\sin^2\beta)\times(1+\sin^4\alpha-2\sin^2\alpha)]+[(1-\sin^2\beta)\times(\sin^4\alpha)]}{(1-\sin^2\beta)(\sin^2\beta)}=1$

$\implies[(\sin^2\beta)\times(1+\sin^4\alpha-2\sin^2\alpha)]+[(1-\sin^2\beta)\times(\sin^4\alpha)]=(1-\sin^2\beta)(\sin^2\beta)$

$\implies(\sin^2\beta+\sin^4\alpha\sin^2\beta-2\sin^2\alpha\sin^2\beta)+(\sin^4\alpha-\sin^4\alpha\sin^2\beta)=\sin^2\beta-\sin^4\beta$

So,

$\implies\sin^2\beta+\sin^4\alpha\sin^2\beta-2\sin^2\alpha\sin^2\beta+\sin^4\alpha-\sin^4\alpha\sin^2\beta=\sin^2\beta-\sin^4\beta$

Cancelling sin²β on both sides,

$\implies\sin^4\alpha\sin^2\beta-2\sin^2\alpha\sin^2\beta+\sin^4\alpha-\sin^4\alpha\sin^2\beta=-\sin^4\beta$

On rearranging,

$\implies(\sin^4\alpha\sin^2\beta-\sin^4\alpha\sin^2\beta)-2\sin^2\alpha\sin^2\beta+\sin^4\alpha=-\sin^4\beta$

On cancelling sin⁴α sin²β,

$\implies-2\sin^2\alpha\sin^2\beta+\sin^4\alpha=-\sin^4\beta$

$\implies\sin^4\alpha+\sin^4\beta-2\sin^2\alpha\sin^2\beta=0$

Transposing 2 sin²α sin²β, to RHS,

$\implies\bf sin^4\alpha+sin^4\beta=2sin^2\alpha sin^2\beta$

And, this is what we had to prove.

HENCE PROVED!!

Answer 2

Answer:

Didi what did Tomboyish told because the Kit is deleting my relevant answers

Step-by-step explanation: