Vardaan Institute - Mathematics Solutions

Topic 1: Finding Principal Value Branches

Ans: $\frac{\pi}{4}$

Ans: $\frac{\pi}{3}$

Ans: $\frac{\pi}{6}$

Ans: $-\frac{\pi}{3}$ (Using $\sin^{-1}(-x) = -\sin^{-1}x$)

Ans: $\frac{3\pi}{4}$ (Using $\cos^{-1}(-x) = \pi - \cos^{-1}x = \pi - \frac{\pi}{4}$)

Ans: $-\frac{\pi}{4}$

10.

Ans: $-\frac{\pi}{4}$

Topic 2: Self-Adjusting & Reciprocal Properties

11.

Ans: $0.5$

12.

Ans: $\frac{\sqrt{3}}{2}$

13.

Ans: $\frac{\pi}{4}$

14.

Ans: $\frac{2\pi}{3}$

15.

Ans: $\frac{\pi}{3}$. Reason: $\frac{2\pi}{3} \notin [-\frac{\pi}{2}, \frac{\pi}{2}]$. Rewrite as $\sin^{-1}(\sin(\pi - \frac{\pi}{3})) = \sin^{-1}(\sin\frac{\pi}{3}) = \frac{\pi}{3}$.

16.

Ans: $\frac{5\pi}{6}$. Reason: Rewrite as $\cos^{-1}(\cos(2\pi - \frac{5\pi}{6})) = \cos^{-1}(\cos\frac{5\pi}{6}) = \frac{5\pi}{6}$.

17.

Ans: $-\frac{\pi}{4}$. Reason: $\tan^{-1}(\tan(\pi - \frac{\pi}{4})) = \tan^{-1}(-\tan\frac{\pi}{4}) = -\frac{\pi}{4}$.

18.

Ans: $\text{cosec}^{-1}(x) = \sin^{-1}(\frac{1}{x})$. Thus, $\text{cosec}^{-1}(2) = \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$.

19.

Ans: $\sec^{-1}(-2) = \cos^{-1}(-\frac{1}{2}) = \pi - \cos^{-1}(\frac{1}{2}) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

20.

Ans: $1$. Reason: $\sin(\frac{\pi}{3} - (-\frac{\pi}{6})) = \sin(\frac{\pi}{3} + \frac{\pi}{6}) = \sin(\frac{\pi}{2}) = 1$.

Topic 3: Even/Odd Nature & Complementary Angles

21.

Ans: $0$. Since $\sin^{-1}(-x) = -\sin^{-1}(x)$.

22.

Ans: $\pi$. Since $\cos^{-1}(-x) + \cos^{-1}(x) = \pi - \cos^{-1}(x) + \cos^{-1}(x) = \pi$.

23.

Ans: $\frac{\pi}{2}$

24.

Ans: $\frac{\pi}{2}$

25.

Ans: $1$. Since $\sin(\frac{\pi}{2}) = 1$.

26.

Ans: $0$. Since $\cos(\frac{\pi}{2}) = 0$.

27.

Ans: $\pi$.

28.

Ans: $\frac{3\pi}{10}$. Reason: $\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x = \frac{\pi}{2} - \frac{\pi}{5} = \frac{3\pi}{10}$.

29.

Ans: $\frac{\pi}{4}$.

30.

Ans: $0$. Since $\cot(\frac{\pi}{2}) = 0$.

Topic 4: Addition & Multiple Angle Formulas

31.

Ans: $\frac{\pi}{4}$. Formula: $\tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right) = \tan^{-1}(1)$.

32.

Ans: $\frac{3\pi}{4}$. Since $xy = 6 > 1$, use $\pi + \tan^{-1}\left(\frac{2+3}{1-6}\right) = \pi + \tan^{-1}(-1) = \pi - \frac{\pi}{4}$.

33.

Ans: $\tan^{-1}\left(\frac{1}{2}\right)$.

34.

Ans: $\tan^{-1}\left(\frac{3}{4}\right)$. Formula: $\tan^{-1}\left(\frac{2(1/3)}{1-(1/9)}\right)$.

35.

Ans: $\sin^{-1}\left(\frac{4}{5}\right)$. Formula: $\sin^{-1}\left(\frac{2(1/2)}{1+(1/4)}\right)$.

36.

Ans: $\cos^{-1}\left(\frac{4}{5}\right)$. Formula: $\cos^{-1}\left(\frac{1-(1/9)}{1+(1/9)}\right)$.

37.

Ans: $\frac{1}{2}$. The inside bracket simplifies to $\tan^{-1}(\frac{1}{2})$.

38.

Ans: $\frac{\pi}{4}$. Divide numerator and denominator of 2nd term by $y$. It becomes $\tan^{-1}(\frac{x}{y}) - [\tan^{-1}(\frac{x}{y}) - \tan^{-1}(1)]$.

39.

Ans: $\pi$. $\frac{\pi}{4} + \frac{3\pi}{4}$ (using the result of Q32).

40.

Ans: $\frac{x+y}{1-xy} = 1 \Rightarrow x + y = 1 - xy$.

Topic 5: Simplification using Trigonometric Substitutions

41.

Ans: $\frac{x}{2}$. Use $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$ and $1+\cos x = 2\cos^2\frac{x}{2}$.

42.

Ans: $\frac{\pi}{4} + \frac{x}{2}$.

43.

Ans: $\frac{x}{2}$. Inner term becomes $\tan(\frac{x}{2})$.

44.

Ans: $\sec^{-1}x$. Substitute $x = \sec\theta$.

45.

Ans: $2\tan^{-1}x$.

46.

Ans: $2\tan^{-1}x$.

47.

Ans: $3\tan^{-1}\left(\frac{x}{a}\right)$. Substitute $x = a\tan\theta$.

48.

Ans: $2\sin^{-1}x$. Substitute $x = \sin\theta$.

49.

Ans: $3\cos^{-1}x$. Substitute $x = \cos\theta$.

50.

Ans: $\frac{1}{2}\tan^{-1}x$. Substitute $x = \tan\theta$.

Topic 6: Solving Inverse Trigonometric Equations

51.

Ans: $x = \frac{1}{2}$.

52.

Ans: $x = -1$.

53.

Ans: $x = \frac{1}{6}$. Reject $x = -1$ as it doesn't satisfy original eq.

54.

Ans: $x = 0$. Standard NCERT problem. $x=1/2$ is an extraneous root.

55.

Ans: $x = \pm\frac{1}{\sqrt{2}}$.

56.

Ans: $x = \frac{\pi}{4}$. Simplifies to $\tan^{-1}(\frac{2\cos x}{1-\cos^2x}) = \tan^{-1}(\frac{2}{\sin x}) \Rightarrow \frac{\cos x}{\sin^2 x} = \frac{1}{\sin x} \Rightarrow \cot x = 1$.

57.

Ans: $x = \pm\frac{3}{4}$. LHS = $\frac{1}{\sqrt{1+x^2}}$, RHS = $\frac{4}{5}$.

58.

Ans: $x = \frac{1}{\sqrt{3}}$. LHS is $\frac{\pi}{4} - \tan^{-1}x$. So, $\frac{\pi}{4} = \frac{3}{2}\tan^{-1}x$.

59.

Ans: $x = \frac{5}{13}$. Compare with $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$.

60.

Ans: $x = \frac{1}{2}\sqrt{\frac{3}{7}}$.

Vardaan Learning Institute