If 2tan^(-1)⁡cos⁡x=tan^(-1)⁡(2cosce⁡x) then sin⁡x+cos⁡x=

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MARKS Content Team · Accepted Answer

Correct answer: (B) √2. As given, 2tan^(-1)⁡ cos⁡ x=tan^(-1)⁡ (2cosce⁡ x) tan^(-1)⁡ cos⁡ x+tan^(-1)⁡ cos⁡ x=tan^(-1)⁡ (2cosce⁡ x) Since cos x ≤ 1 =tan^(-1)⁡ (2cos⁡ x)/(1-cos^2⁡ x)=tan^(-1)⁡ (2cosce⁡ x) ⇒ (2cos⁡ x)/(sin^2⁡ x)=2cosce⁡ x ⇒ 2cot⁡ x=2 ⇒ cot⁡ x=1 ⇒ x=(π)/4 Hence, sin⁡ x+cos⁡ x=sin⁡(π)/4+cos⁡(π)/4 =1/(√2)+1/(√2) =√2

If $2 \tan^{- 1} (\cos x) = \tan^{- 1} (2 cosce x) t h e n \sin x + \cos x =$

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If 2tan-1⁡cos⁡x=tan-1⁡(2cosce⁡x) then sin⁡x+cos⁡x=

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If $2 \tan^{- 1} (\cos x) = \tan^{- 1} (2 cosce x) t h e n \sin x + \cos x =$