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Example 5 - Express tan-1 cos⁡x/(1 - sin⁡x) - Chapter 2 Inverse

Example 5 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 2
Example 5 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 3 Example 5 - Chapter 2 Class 12 Inverse Trigonometric Functions - Part 4

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Example 5 Express tan−1 cos⁡x/(1 − sin⁡x ) , – π/2 < x < 3π/2 in the simplest form Lets first calculate cos x & 1 – sin x We know that cos 2x = 𝐜𝐨𝐬𝟐⁡𝐱 – 𝐬𝐢𝐧𝟐⁡𝐱 Replacing x by 𝑥/2 cos (2x/2) = cos2 x/2 – sin2 x/2 cos x = cos2 x/2 – sin2 x/2 We know that sin 2x = 2 sin x cos x Replacing x by 𝑥/2 sin (2𝑥/2) = 2 sin 𝑥/2 cos 𝑥/2 sin x = 2 sin 𝑥/2 cos 𝑥/2 Solving tan−1 (cos⁡x/(1 〖− sin〗⁡x )) = tan−1 [(cos2 x/2 − sin2 x/2)/(1 − (2 〖sin 〗⁡〖x/2 cos⁡〖 x/2〗 〗 ) )] = tan−1 [(cos2 x/2 − sin2 x/2)/(1 − 2 〖sin 〗⁡〖x/2 〖 cos 〗⁡〖x/2〗 〗 )] = tan−1 [(cos2 x/2 − sin2 x/2)/(cos2 x/2 + sin2 x/2 − 2 〖sin 〗⁡〖x/2 cos⁡〖 x/2〗 〗 )] As sin2 x + cos2 x = 1 Replacing x by 𝑥/2 sin2 𝑥/2 + cos2 𝑥/2 = 1 1 = sin2 𝑥/2 + cos2 𝑥/2 = tan−1 [(cos x/2 + sin x/2)(cos x/2 − sin x/2)/(cos x/2 − sin x/2)^2 ] = tan−1 [((cos x/2 + sin x/2))/((cos x/2 − sin x/2) )] Dividing by cos 𝑥/2 = tan−1 ((cos⁡〖 𝑥/( 2 )〗/〖𝑐𝑜𝑠 〗⁡〖 𝑥/2〗 + sin⁡〖 𝑥/( 2 )〗/〖𝑐𝑜𝑠 〗⁡〖 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 [(𝟏 + 〖tan 〗⁡〖𝑥/2〗)/(1 − 〖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 Using tan (x + y ) = 𝒕𝒂𝒏⁡〖𝒙 +〖 𝒕𝒂𝒏〗⁡〖𝒚 〗 〗/(𝟏 − 𝒕𝒂𝒏⁡〖𝒙 𝒕𝒂𝒏⁡𝒚 〗 ) Replace x by 𝜋/4 and y by 𝑥/2

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.