CBSE Class 12 Mathematics Inverse Trignometric Function Worksheet ... - Free Printable

Here are the step-by-step solutions for the problems in the image.

1. Find the principal value of the following:

a) $\sin^{-1}(\frac{1}{\sqrt{2}})$
* Let $y = \sin^{-1}(\frac{1}{\sqrt{2}})$. Then $\sin y = \frac{1}{\sqrt{2}}$.
* The range of principal values for sine is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
* We know that $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
* Answer: $\frac{\pi}{4}$

b) $\cos^{-1}(-\frac{1}{\sqrt{2}})$
* Let $y = \cos^{-1}(-\frac{1}{\sqrt{2}})$. Then $\cos y = -\frac{1}{\sqrt{2}}$.
* The range for cosine is $[0, \pi]$. Since the value is negative, the angle is in the second quadrant.
* $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$, so $\cos(\pi - \frac{\pi}{4}) = -\frac{1}{\sqrt{2}}$.
* $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
* Answer: $\frac{3\pi}{4}$

c) $\tan^{-1}(\frac{-1}{\sqrt{3}})$
* Let $y = \tan^{-1}(\frac{-1}{\sqrt{3}})$. Then $\tan y = -\frac{1}{\sqrt{3}}$.
* The range for tangent is $(-\frac{\pi}{2}, \frac{\pi}{2})$.
* We know $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$. Since it's negative, we take the negative angle.
* Answer: $-\frac{\pi}{6}$

d) $\text{cosec}^{-1}(-2)$
* Let $y = \text{cosec}^{-1}(-2)$. This means $\text{cosec } y = -2$, or $\sin y = -\frac{1}{2}$.
* The range for cosec is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ excluding 0.
* $\sin(\frac{\pi}{6}) = \frac{1}{2}$, so $\sin(-\frac{\pi}{6}) = -\frac{1}{2}$.
* Answer: $-\frac{\pi}{6}$

e) $\sec^{-1}(-\frac{2}{\sqrt{3}})$
* Let $y = \sec^{-1}(-\frac{2}{\sqrt{3}})$. This means $\sec y = -\frac{2}{\sqrt{3}}$, or $\cos y = -\frac{\sqrt{3}}{2}$.
* The range for sec is $[0, \pi]$ excluding $\frac{\pi}{2}$.
* $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$. In the second quadrant, the angle is $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
* Answer: $\frac{5\pi}{6}$

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2. Find the value of the following:

a) $\sin^{-1}(\sin \frac{2\pi}{3})$
* The property $\sin^{-1}(\sin x) = x$ only works if $x$ is between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.
* $\frac{2\pi}{3}$ is outside this range.
* $\sin(\frac{2\pi}{3}) = \sin(\pi - \frac{\pi}{3}) = \sin(\frac{\pi}{3})$.
* Now, $\sin^{-1}(\sin \frac{\pi}{3}) = \frac{\pi}{3}$ (which is inside the range).
* Answer: $\frac{\pi}{3}$

b) $\cos^{-1}(\cos \frac{13\pi}{6})$
* The range for $\cos^{-1}$ is $[0, \pi]$.
* $\frac{13\pi}{6} = 2\pi + \frac{\pi}{6}$.
* $\cos(2\pi + \frac{\pi}{6}) = \cos(\frac{\pi}{6})$.
* So, $\cos^{-1}(\cos \frac{\pi}{6}) = \frac{\pi}{6}$.
* Answer: $\frac{\pi}{6}$

c) $\tan^{-1}(\tan \frac{7\pi}{6})$
* The range for $\tan^{-1}$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.
* $\frac{7\pi}{6} = \pi + \frac{\pi}{6}$.
* $\tan(\pi + \frac{\pi}{6}) = \tan(\frac{\pi}{6})$.
* So, $\tan^{-1}(\tan \frac{\pi}{6}) = \frac{\pi}{6}$.
* Answer: $\frac{\pi}{6}$

d) $\text{cosec}^{-1}(\text{cosec } \frac{\pi}{8})$
* The range for $\text{cosec}^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$ excluding 0.
* $\frac{\pi}{8}$ is within this range.
* Therefore, the answer is simply the angle itself.
* Answer: $\frac{\pi}{8}$

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3. Evaluate the following:

a) $\sin \{ \frac{\pi}{3} - \sin^{-1}(-\frac{1}{2}) \}$
* First, find $\sin^{-1}(-\frac{1}{2})$. We know $\sin(-\frac{\pi}{6}) = -\frac{1}{2}$, so the value is $-\frac{\pi}{6}$.
* Substitute this back: $\sin \{ \frac{\pi}{3} - (-\frac{\pi}{6}) \} = \sin \{ \frac{\pi}{3} + \frac{\pi}{6} \}$.
* $\frac{\pi}{3} + \frac{\pi}{6} = \frac{2\pi}{6} + \frac{\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$.
* $\sin(\frac{\pi}{2}) = 1$.
* Answer: $1$

b) $\sin (\frac{1}{2} \cos^{-1} \frac{4}{5})$
* Let $\theta = \cos^{-1} \frac{4}{5}$. Then $\cos \theta = \frac{4}{5}$.
* We need to find $\sin(\frac{\theta}{2})$.
* Formula: $\sin(\frac{\theta}{2}) = \sqrt{\frac{1 - \cos \theta}{2}}$.
* $\sin(\frac{\theta}{2}) = \sqrt{\frac{1 - 4/5}{2}} = \sqrt{\frac{1/5}{2}} = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}}$.
* Answer: $\frac{1}{\sqrt{10}}$

c) $\tan \frac{1}{2} (\cos^{-1} \frac{\sqrt{5}}{3})$
* Let $\theta = \cos^{-1} \frac{\sqrt{5}}{3}$. Then $\cos \theta = \frac{\sqrt{5}}{3}$.
* We need $\tan(\frac{\theta}{2})$.
* Formula: $\tan(\frac{\theta}{2}) = \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}$.
* $\tan(\frac{\theta}{2}) = \sqrt{\frac{1 - \frac{\sqrt{5}}{3}}{1 + \frac{\sqrt{5}}{3}}} = \sqrt{\frac{\frac{3-\sqrt{5}}{3}}{\frac{3+\sqrt{5}}{3}}} = \sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}$.
* Rationalize the denominator inside the square root by multiplying top and bottom by $(3-\sqrt{5})$:
* Numerator: $(3-\sqrt{5})^2$
* Denominator: $3^2 - (\sqrt{5})^2 = 9 - 5 = 4$
* Expression becomes $\sqrt{\frac{(3-\sqrt{5})^2}{4}} = \frac{3-\sqrt{5}}{2}$.
* Answer: $\frac{3-\sqrt{5}}{2}$

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4. Evaluate: $\cos (\sin^{-1} \frac{1}{5} + \cos^{-1} \frac{12}{13})$

* Let $A = \sin^{-1} \frac{1}{5}$ and $B = \cos^{-1} \frac{12}{13}$.
* We need to find $\cos(A+B) = \cos A \cos B - \sin A \sin B$.
* From $A = \sin^{-1} \frac{1}{5}$: $\sin A = \frac{1}{5}$. Using Pythagoras ($1^2 + b^2 = 5^2$), adjacent side is $\sqrt{24} = 2\sqrt{6}$. So $\cos A = \frac{2\sqrt{6}}{5}$.
* From $B = \cos^{-1} \frac{12}{13}$: $\cos B = \frac{12}{13}$. Using Pythagoras ($12^2 + a^2 = 13^2$), opposite side is $5$. So $\sin B = \frac{5}{13}$.
* Substitute into formula:
$(\frac{2\sqrt{6}}{5})(\frac{12}{13}) - (\frac{1}{5})(\frac{5}{13})$
$= \frac{24\sqrt{6}}{65} - \frac{5}{65}$
$= \frac{24\sqrt{6} - 5}{65}$
* Answer: $\frac{24\sqrt{6} - 5}{65}$

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5. Show that: $\tan^{-1}(\sqrt{x}) = \frac{1}{2} \cos^{-1}(\frac{1-x}{1+x})$

* Let $\sqrt{x} = \tan \theta$. Then $x = \tan^2 \theta$ and $\theta = \tan^{-1}(\sqrt{x})$.
* Consider the Right Hand Side (RHS): $\frac{1}{2} \cos^{-1}(\frac{1-\tan^2 \theta}{1+\tan^2 \theta})$.
* We know the identity $\cos 2\theta = \frac{1-\tan^2 \theta}{1+\tan^2 \theta}$.
* So, RHS $= \frac{1}{2} \cos^{-1}(\cos 2\theta)$.
* RHS $= \frac{1}{2} (2\theta) = \theta$.
* Since $\theta = \tan^{-1}(\sqrt{x})$, LHS = RHS.
* Proved.

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6. Prove that $\tan^{-1}(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}) = \frac{\pi}{4} + \frac{1}{2}\cos^{-1}x^2$

* Let $x^2 = \cos 2\theta$. Then $\theta = \frac{1}{2}\cos^{-1}x^2$.
* Substitute $x^2$ in the expression:
$\sqrt{1+x^2} = \sqrt{1+\cos 2\theta} = \sqrt{2\cos^2 \theta} = \sqrt{2}\cos \theta$
$\sqrt{1-x^2} = \sqrt{1-\cos 2\theta} = \sqrt{2\sin^2 \theta} = \sqrt{2}\sin \theta$
* The fraction becomes:
$\frac{\sqrt{2}\cos \theta - \sqrt{2}\sin \theta}{\sqrt{2}\cos \theta + \sqrt{2}\sin \theta} = \frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}$
* Divide numerator and denominator by $\cos \theta$:
$\frac{1 - \tan \theta}{1 + \tan \theta} = \tan(\frac{\pi}{4} - \theta)$
* So, LHS $= \tan^{-1}(\tan(\frac{\pi}{4} - \theta)) = \frac{\pi}{4} - \theta$.
* Wait, the question asks to prove it equals $\frac{\pi}{4} + \frac{1}{2}\cos^{-1}x^2$. Let's re-check the standard identity. Usually, this specific form results in $\frac{\pi}{4} - \frac{1}{2}\cos^{-1}x^2$.
* Let's check the sign in the question image carefully. It says $\frac{\pi}{4} + \frac{1}{2}\cos^{-1}x^2$.
* If we assume the question meant $\tan^{-1}(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}})$, it would be plus. Or if the numerator was swapped.
* However, based on standard derivation: $\tan^{-1}(\frac{1-\tan \alpha}{1+\tan \alpha}) = \frac{\pi}{4} - \alpha$. Here $\alpha = \theta$. So result is $\frac{\pi}{4} - \theta$.
* There might be a typo in the question paper (common in these sets) where it should be minus, or the terms in the fraction are swapped. Assuming standard identities, the derivation leads to $\frac{\pi}{4} - \frac{1}{2}\cos^{-1}x^2$.
* *Correction*: If we let $x^2 = \cos 2\theta$, then $\theta \in [0, \pi/2]$. The term $\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\dots}$ simplifies to $\tan(\frac{\pi}{4}-\theta)$. Thus LHS $= \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x^2$.
* If the question strictly requires proving the "+" version, the numerator/denominator in the image might be interpreted differently or there is a typo in the source. I will provide the standard proof which yields the minus sign, as that is mathematically correct for the written expression.

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7. Prove that $\tan^{-1} \frac{1}{4} + \tan^{-1} \frac{2}{9} = \frac{1}{2} \cos^{-1} \frac{3}{5}$

* LHS: Use formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \frac{x+y}{1-xy}$.
$x=\frac{1}{4}, y=\frac{2}{9}$.
Numerator: $\frac{1}{4} + \frac{2}{9} = \frac{9+8}{36} = \frac{17}{36}$.
Denominator: $1 - (\frac{1}{4})(\frac{2}{9}) = 1 - \frac{2}{36} = \frac{34}{36}$.
Fraction: $\frac{17/36}{34/36} = \frac{17}{34} = \frac{1}{2}$.
So, LHS $= \tan^{-1} \frac{1}{2}$.
* RHS: Let $\alpha = \cos^{-1} \frac{3}{5}$. Then $\cos \alpha = \frac{3}{5}$.
Draw a triangle: Adjacent=3, Hypotenuse=5. Opposite=$\sqrt{25-9}=4$.
So $\tan \alpha = \frac{4}{3}$.
We need $\frac{1}{2} \alpha$. Let $\beta = \frac{\alpha}{2}$.
$\tan \alpha = \frac{2 \tan \beta}{1 - \tan^2 \beta} = \frac{4}{3}$.
$3(2t) = 4(1-t^2) \Rightarrow 6t = 4 - 4t^2 \Rightarrow 4t^2 + 6t - 4 = 0 \Rightarrow 2t^2 + 3t - 2 = 0$.
$(2t-1)(t+2)=0$. So $t = \frac{1}{2}$ or $-2$. Since angle is acute, $\tan \beta = \frac{1}{2}$.
So RHS $= \tan^{-1} \frac{1}{2}$.
* LHS = RHS. Proved.

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8. Prove that $\cot^{-1}(\frac{ab+1}{a-b}) + \cot^{-1}(\frac{bc+1}{b-c}) + \cot^{-1}(\frac{ca+1}{c-a}) = 0$

* Recall $\cot^{-1} x = \tan^{-1} \frac{1}{x}$.
* Term 1: $\tan^{-1} \frac{a-b}{1+ab} = \tan^{-1} a - \tan^{-1} b$.
* Term 2: $\tan^{-1} \frac{b-c}{1+bc} = \tan^{-1} b - \tan^{-1} c$.
* Term 3: $\tan^{-1} \frac{c-a}{1+ca} = \tan^{-1} c - \tan^{-1} a$.
* Sum: $(\tan^{-1} a - \tan^{-1} b) + (\tan^{-1} b - \tan^{-1} c) + (\tan^{-1} c - \tan^{-1} a)$.
* Everything cancels out.
* Result = 0. Proved.

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9. Prove that $\tan^{-1} \frac{1}{4} + \tan^{-1} \frac{3}{5} - \tan^{-1} \frac{8}{19} = \frac{\pi}{4}$

* Combine first two terms: $\tan^{-1} \frac{1/4 + 3/5}{1 - (1/4)(3/5)} = \tan^{-1} \frac{17/20}{17/20} = \tan^{-1}(1) = \frac{\pi}{4}$.
* So the expression becomes $\frac{\pi}{4} - \tan^{-1} \frac{8}{19}$.
* This does not equal $\frac{\pi}{4}$ unless $\tan^{-1} \frac{8}{19} = 0$, which is false.
* *Re-evaluating*: Did I copy the numbers right? Image says $\tan^{-1} \frac{1}{4} + \tan^{-1} \frac{3}{5} - \tan^{-1} \frac{8}{19} = \frac{\pi}{4}$.
Let's check $\tan^{-1} \frac{1}{4} + \tan^{-1} \frac{3}{5}$.
Num: $5+12=17$. Denom: $20-3=17$. Ratio 1. Angle $\pi/4$.
So LHS is $\frac{\pi}{4} - \tan^{-1} \frac{8}{19}$.
This implies the question statement in the image might have a typo (perhaps the last term shouldn't be there, or the target is different).
*Alternative interpretation*: Maybe the last term is added? $\frac{\pi}{4} + \tan^{-1} \frac{8}{19} \neq \frac{\pi}{4}$.
Actually, looking at similar textbook problems, often the sum is $\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3} = \frac{\pi}{4}$.
With the given numbers, the LHS evaluates to approx $0.785 - 0.398 \neq 0.785$.
Note: Based strictly on calculation, the equation as written is incorrect. However, if the question intended $\tan^{-1} \frac{1}{4} + \tan^{-1} \frac{3}{5} = \frac{\pi}{4}$, that part is true. The subtraction of the third term makes it false. I will assume there is a typo in the question paper provided in the image.

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10. Prove that $\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}$

* Let $A = \cos^{-1} \frac{4}{5}$ and $B = \cos^{-1} \frac{12}{13}$.
* $\cos A = \frac{4}{5} \Rightarrow \sin A = \frac{3}{5}$.
* $\cos B = \frac{12}{13} \Rightarrow \sin B = \frac{5}{13}$.
* We want $\cos(A+B) = \cos A \cos B - \sin A \sin B$.
* $= (\frac{4}{5})(\frac{12}{13}) - (\frac{3}{5})(\frac{5}{13})$
* $= \frac{48}{65} - \frac{15}{65} = \frac{33}{65}$.
* So $A+B = \cos^{-1} \frac{33}{65}$. Proved.

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11. Solve for $x$: $\sin^{-1}(1-x) - 2\sin^{-1} x = \frac{\pi}{2}$

* Rearrange: $\sin^{-1}(1-x) = \frac{\pi}{2} + 2\sin^{-1} x$.
* Take sine of both sides:
$1-x = \sin(\frac{\pi}{2} + 2\sin^{-1} x)$.
* Identity: $\sin(\frac{\pi}{2} + \theta) = \cos \theta$.
So, $1-x = \cos(2\sin^{-1} x)$.
* Let $\alpha = \sin^{-1} x$. Then $\sin \alpha = x$.
$\cos 2\alpha = 1 - 2\sin^2 \alpha = 1 - 2x^2$.
* Equation: $1-x = 1 - 2x^2$.
* $2x^2 - x = 0 \Rightarrow x(2x-1) = 0$.
* Possible solutions: $x=0$ or $x=\frac{1}{2}$.
* Check $x=0$: $\sin^{-1}(1) - 0 = \frac{\pi}{2}$. Correct.
* Check $x=\frac{1}{2}$: $\sin^{-1}(\frac{1}{2}) - 2\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6} - 2(\frac{\pi}{6}) = -\frac{\pi}{6} \neq \frac{\pi}{2}$. Incorrect.
* Answer: $x = 0$

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12. If $\cos^{-1} \frac{x}{a} + \cos^{-1} \frac{y}{b} = \theta$, prove that $\frac{x^2}{a^2} - \frac{2xy}{ab} \cos \theta + \frac{y^2}{b^2} = \sin^2 \theta$

* Let $A = \cos^{-1} \frac{x}{a}$ and $B = \cos^{-1} \frac{y}{b}$.
* So $\cos A = \frac{x}{a}$ and $\cos B = \frac{y}{b}$.
* Given $A+B = \theta$.
* $\cos(A+B) = \cos \theta$.
* Expand: $\cos A \cos B - \sin A \sin B = \cos \theta$.
* $\frac{x}{a} \frac{y}{b} - \sqrt{1-\frac{x^2}{a^2}} \sqrt{1-\frac{y^2}{b^2}} = \cos \theta$.
* Rearrange: $\frac{xy}{ab} - \cos \theta = \sqrt{1-\frac{x^2}{a^2}} \sqrt{1-\frac{y^2}{b^2}}$.
* Square both sides:
$(\frac{xy}{ab} - \cos \theta)^2 = (1-\frac{x^2}{a^2})(1-\frac{y^2}{b^2})$.
* LHS: $\frac{x^2 y^2}{a^2 b^2} - \frac{2xy}{ab} \cos \theta + \cos^2 \theta$.
* RHS: $1 - \frac{y^2}{b^2} - \frac{x^2}{a^2} + \frac{x^2 y^2}{a^2 b^2}$.
* Cancel $\frac{x^2 y^2}{a^2 b^2}$ from both sides.
* $- \frac{2xy}{ab} \cos \theta + \cos^2 \theta = 1 - \frac{y^2}{b^2} - \frac{x^2}{a^2}$.
* Move terms to left: $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab} \cos \theta = 1 - \cos^2 \theta$.
* Since $1 - \cos^2 \theta = \sin^2 \theta$, we get:
$\frac{x^2}{a^2} - \frac{2xy}{ab} \cos \theta + \frac{y^2}{b^2} = \sin^2 \theta$. Proved.

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13. Prove that $\tan(\frac{\pi}{4} + \frac{1}{2} \cos^{-1} \frac{a}{b}) + \tan(\frac{\pi}{4} - \frac{1}{2} \cos^{-1} \frac{a}{b}) = \frac{2b}{a}$

* Let $\alpha = \frac{1}{2} \cos^{-1} \frac{a}{b}$. Then $\cos 2\alpha = \frac{a}{b}$.
* We need to evaluate $\tan(\frac{\pi}{4} + \alpha) + \tan(\frac{\pi}{4} - \alpha)$.
* Using expansion: $\frac{1+\tan \alpha}{1-\tan \alpha} + \frac{1-\tan \alpha}{1+\tan \alpha}$.
* Common denominator: $1 - \tan^2 \alpha$.
* Numerator: $(1+\tan \alpha)^2 + (1-\tan \alpha)^2 = 1 + 2\tan \alpha + \tan^2 \alpha + 1 - 2\tan \alpha + \tan^2 \alpha = 2(1+\tan^2 \alpha) = 2\sec^2 \alpha$.
* Expression: $\frac{2\sec^2 \alpha}{1-\tan^2 \alpha}$.
* Multiply top and bottom by $\cos^2 \alpha$:
$\frac{2}{\cos^2 \alpha - \sin^2 \alpha} = \frac{2}{\cos 2\alpha}$.
* Since $\cos 2\alpha = \frac{a}{b}$, the expression is $\frac{2}{a/b} = \frac{2b}{a}$. Proved.

Final Answer:
See the step-by-step proofs and evaluations above for questions 1 through 13.