Example 10 - Chapter 2 Class 12 Inverse Trigonometric Functions

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Example 10 Show that sin−1 3/5 − sin-1 8/17 = cos−1 84/85 Let sin−1 3/5 = a and sin−1 8/17 = b Let sin−1 𝟑/𝟓 = a sin a = 3/5 cos a = √(1−sin2 a) = √(1 −(3/5)^2 ) = √(1 − 9/25) = 4/5 cos a = 4/5 Let sin−1 𝟖/𝟏𝟕 = b sin b = 8/17 cos b = √(1 −sin2 b) = √(1 −(8/17)^2 ) = √(1 −64/289) = 15/17 cos b = 15/17 We convert sin-1 to cos-1 & then use cos (a – b) formula We know that cos (a – b) = cos a cos b + sin a sin b cos (a – b) = 4/5 × 15/17 + 3/5 × 8/17 cos (a – b) = (60 + 24)/(17 × 5) cos (a – b) = 84/85 (a – b) = cos−1 (84/85) Putting the value of a and b Putting cos a = 4/5 & sin a = 3/5 and cos b = 15/13 & sin b = 8/17 sin−1 3/5 – sin−1 8/7 = cos−1 (84/85) Hence proved