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Chapter 2 Class 12 Inverse Trigonometric Functions
Ex 2.1, 8 Important
Ex 2.1, 12 Important
Ex 2.1, 14 (MCQ) Important
Example 4 Important You are here
Question 3 Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Ex 2.2, 13 (MCQ) Important
Ex 2.2, 15 (MCQ)
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Misc 2 Important
Misc 7 Important
Misc 9 Important
Misc 10 Important
Question 2 Important Deleted for CBSE Board 2024 Exams
Question 3 (MCQ) Deleted for CBSE Board 2024 Exams
Chapter 2 Class 12 Inverse Trigonometric Functions
Last updated at May 29, 2023 by Teachoo
Example 4 Express tan−1 cosx/(1 − sinx ) , – π/2 < x < 3π/2 in the simplest form Lets first calculate cos x & 1 – sin x We know that cos 2x = 𝐜𝐨𝐬𝟐𝐱 – 𝐬𝐢𝐧𝟐𝐱 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 (cosx/(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 𝒕𝒂𝒏𝒚 〗 ) Replace x by 𝜋/4 and y by 𝑥/2