Simplest form of tan−1 ((√(1 + cos ⁡x )   + √(1 - cos⁡x ))/(√(1 + cos⁡x )   - √(1  - cos⁡x ))), π < x < 3π/2 is :

(a) π/4 − x/2                (b) 3π/2 − x/2   
(c) − x/2                       (d) π − x/2

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Question 27 Simplest form of tan−1 ((√(1 +〖 cos〗⁡𝑥 ) + √(1 − cos⁡𝑥 ))/(√(1 + cos⁡𝑥 ) − √(1 − cos⁡𝑥 ))), 𝜋 < x < 3𝜋/2 is : (a) 𝜋/4 − 𝑥/2 (b) 3𝜋/2 − 𝑥/2 (c) − 𝑥/2 (d) 𝜋 − 𝑥/2 We know that cos 2x = 2 cos2 x − 1 Replace x by 𝑥/2 cos x = 2 cos2 𝑥/2 − 1 Adding 1 both sides 1 + cos x = 2 cos2 𝑥/2 √(𝟏+𝒄𝒐𝒔⁡𝒙 ) = √𝟐 cos 𝒙/𝟐 We know that cos 2x = 1 − 2 sin2 x 1 − cos 2x = 2 sin2 x Replace x by 𝑥/2 1 − cos x = 2 sin2 𝑥/2 √(𝟏−𝒄𝒐𝒔⁡𝒙 ) = √𝟐 sin 𝒙/𝟐 Therefore, tan−1 ((√(1 +〖 cos〗⁡𝑥 ) + √(1 − cos⁡𝑥 ))/(√(1 + cos⁡𝑥 ) − √(1 − cos⁡𝑥 ))) = tan−1 ((√𝟐 〖𝒄𝒐𝒔 〗⁡〖𝒙/𝟐〗 + √𝟐 〖𝒔𝒊𝒏 〗⁡〖𝒙/𝟐〗 )/(√𝟐 〖𝒄𝒐𝒔 〗⁡〖𝒙/𝟐〗 − √𝟐 〖𝒔𝒊𝒏 〗⁡〖𝒙/𝟐〗 )) = tan−1 ((〖cos 〗⁡〖𝑥/2〗 − 〖sin 〗⁡〖𝑥/2〗 )/(〖cos 〗⁡〖𝑥/2〗 + 〖sin 〗⁡〖𝑥/2〗 )) Dividing by 〖𝒄𝒐𝒔 〗⁡〖𝒙/𝟐〗 inside = tan−1 (((〖cos 〗⁡〖𝑥/2〗 − 〖sin 〗⁡〖𝑥/2〗)/(cos⁡𝑥/2))/((〖cos 〗⁡〖𝑥/2〗 + 〖sin 〗⁡〖𝑥/2〗)/(cos⁡𝑥/2))) = tan−1 ((1 − tan⁡〖 x/2〗)/(1 +〖 tan〗⁡〖 x/2〗 )) = tan−1 ((𝒕𝒂𝒏⁡〖 𝝅/𝟒〗 − tan⁡〖 𝑥/2〗)/(1 + 〖𝐭𝐚𝐧 〗⁡〖𝝅/𝟒 .〖 tan 〗⁡〖𝑥/2〗 〗 )) = tan−1 ("tan " (𝜋/4−𝑥/2)) = 𝝅/𝟒 − 𝒙/𝟐 So, the correct answer is (A)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo