Chapter 2 Class 12 Inverse Trigonometric Functions

Serial order wise

Last updated at April 16, 2024 by Teachoo

Question 6 (Introduction) Simplify tan−1 [(a cos〖x − b sinx 〗)/(b cos〖x + a sinx 〗 )], if a/b tan x > −1 We write (a cos〖x − b sinx 〗)/(b cos〖x + a sinx 〗 ) in form of tan We know that tan (x – y) = 𝑡𝑎𝑛〖𝑥 −〖 𝑡𝑎𝑛〗〖𝑦 〗 〗/(1 + 𝑡𝑎𝑛〖𝑥 𝑡𝑎𝑛𝑦 〗 ) We need denominator in form 1 + tan x tan y Hence, we need 1 instead of b cos x So dividing both numerator and denominator by b cos x Question 6 Simplify tan−1 [(a cos〖x − b sinx 〗)/(b cos〖x + a sinx 〗 )], if a/b tan x > −1 tan−1 [(a cos〖x − b sinx 〗)/(b cos〖x + a sinx 〗 )] = tan−1 [((a cos〖x − b sinx 〗)/(b cosx ))/((b cos〖x + a sinx 〗)/(b cosx ))] = tan−1 [((𝑎 cos𝑥)/(𝑏 cos𝑥 ) − (𝑏 sin𝑥)/(𝑏 cos𝑥 ))/((𝑏 cos𝑥)/(𝑏 cos𝑥 ) + (𝑎 sin𝑥)/(𝑏 cos𝑥 ))] = tan−1 [(𝑎/(𝑏 ) − sin𝑥/cos𝑥 )/(1 + (𝑎 sin𝑥)/(𝑏 cos𝑥 ))] = tan−1 [(a/b − tanx)/(1 + a/b tanx )] = tan−1 a/b – tan−1 (tan x) = tan−1 𝐚/𝐛 − x Using equation tan−1((𝒙 − 𝒚)/(𝟏 + 𝒙𝒚)) = tan−1 x – tan−1 y Replacing x with 𝑎/𝑏 and y with tan x