Question

please guys solve both​

Answer 1

Answer:

Unable to see the question

Answer 2

Solution:

(i) To prove: $\sqrt{2+\sqrt{2+2cos4\beta}} = 2cos\beta$

Taking L.H.S.,

= $\sqrt{2+\sqrt{2+2(2cos^{2}2\beta-1)}}$            {∵cos2θ = 2cos²θ-1}

= $\sqrt{2+\sqrt{2+4cos^{2}2\beta-2}}$

= $\sqrt{2+\sqrt{4cos^{2}2\beta}}$

= $\sqrt{2+2cos2\beta}$

= $\sqrt{2(1+cos2\beta)}$

= $\sqrt{2*2cos^{2}\beta}$

= $\sqrt{4cos^{2}\beta}$

= 2cosβ = R.H.S.

Hence Proved

(ii) I think, your question is :- To prove: $\sqrt{2+\sqrt{2+\sqrt{2+2cos8\beta}}} = 2cos\beta$

Taking L.H.S.,

= $\sqrt{2+\sqrt{2+\sqrt{2+2cos8\beta}}}$

= $\sqrt{2+\sqrt{2+\sqrt{2+4cos^{2}4\beta-2}}}$

= $\sqrt{2+\sqrt{2+\sqrt{4cos^{2}4\beta}}}$

= $\sqrt{2+\sqrt{2+2cos4\beta}}$

= $\sqrt{2+\sqrt{2(1+cos4\beta)}}$

= $\sqrt{2+\sqrt{2*2cos^{2}2\beta}}$

= $\sqrt{2+\sqrt{4cos^{2}2\beta}}$

= $\sqrt{2+2cos2\beta}$

= $\sqrt{2(1+cos2\beta)}$

= $\sqrt{2*2cos^{2}\beta}$

= $\sqrt{4cos^{2}\beta}$

= 2cosβ = R.H.S.

Hence Proved

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