If sin θ > cos θ, then the value of \({ \sqrt{1-2\sin\theta \cos\theta} + \sqrt{1+2\sin\theta \cos\theta} \over 2\tan\theta}\) is:

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RRB Group D 29 Aug 2022 Shift 3 Official Paper

Answer 1

Given:

\({ \sqrt{1-2\sin\theta \cos\theta} + \sqrt{1+2\sin\theta \cos\theta} \over 2\tan\theta}\)

Formula used:

\(sin^2\theta + cos^2\theta=1\)

\(a^2 + b^2 +2ab =(a+b)^2\)

\(a^2 + b^2 -2ab =(a-b)^2\)

Calculation:

\(\sqrt{1-2\sin\theta \cos\theta}\)

=> \(\sqrt{sin^2\theta+cos^2\theta-2\sin\theta \cos\theta}\)

=> \(\sqrt{(sin\theta- \cos\theta)^2 \:}\)

=> \((sin\theta -cos\theta)\)

Similarly,

\(\sqrt{1+2\sin\theta \cos\theta}\) = \((sin\theta +cos\theta)\)

now put the values in given equation,

\({ \sqrt{1-2\sin\theta \cos\theta} + \sqrt{1+2\sin\theta \cos\theta} \over 2\tan\theta}\)

=> \({{ (sin\theta-cos\theta)} + (sin\theta+cos\theta) \over 2\tan\theta}\)

=> \(2\sin\theta\over{2{sin\theta\over cos\theta}}\)

=> \(cos\theta\)

Hence, the correct answer is 'Cos θ'.

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