Exercise 9.2 Class 12 Math Solutions - Unit 9 Inverse Trigonometric Functions, Punjab Board

Exercise 9.2 of Unit 9: Inverse Trigonometric Functions and Their Graphs in Class 12 Mathematics under the Punjab Board PCTB / PECTAA curriculum focuses on algebraic addition and subtraction identities for inverse circular functions. By defining angles (\alpha = \sin^{-1} x) ((\sin\alpha = x, \cos\alpha = \sqrt{1 - x^2})) and (\beta = \sin^{-1} y) ((\sin\beta = y, \cos\beta = \sqrt{1 - y^2})), the trigonometric compound identity (\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta) establishes the inverse sine theorem: (\sin^{-1} x \pm \sin^{-1} y = \sin^{-1}\left(x\sqrt{1 - y^2} \pm y\sqrt{1 - x^2}\right)).

Similarly, using (\cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta) yields the inverse cosine theorem: (\cos^{-1} x \pm \cos^{-1} y = \cos^{-1}\left(xy \mp \sqrt{1 - x^2}\sqrt{1 - y^2}\right)). For inverse tangents, (\tan(\alpha \pm \beta) = \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha\tan\beta}) produces (\tan^{-1} x \pm \tan^{-1} y = \tan^{-1}\left(\frac{x \pm y}{1 \mp xy}\right)) (valid when (xy < 1)). These identities enable students to prove double-angle formulas (2\tan^{-1} x = \tan^{-1}\left(\frac{2x}{1 - x^2}\right) = \sin^{-1}\left(\frac{2x}{1 + x^2}\right) = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right)) and simplify sums of multiple inverse tangent terms.

For Punjab Board HSSC Part 2 examinations across all nine educational divisions (Lahore, Rawalpindi, Gujranwala, Faisalabad, Multan, Bahawalpur, Sargodha, Sahiwal, and D.G. Khan), Exercise 9.2 is a prime source of Section B Short Questions (evaluating numerical inverse sums such as (\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \frac{\pi}{4})) and Section C Long Questions (proving multi-step identities involving mixed inverse functions). These step-by-step solutions provide rigorous variable substitutions, clear trigonometric reductions, and complete algebraic justifications for every textbook problem.