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Vidaara.orgClass 12 · Mathematics
CodeVID-M12-02-CT
Inverse Trigonometric Functions — Full Chapter Test
Chapter: Inverse Trigonometric Functions
Topic: Complete chapter — all topics
Maximum Marks: 35
Time: 80 minutes
Name: ____________________ Roll No.: __________ Date: ____________

General Instructions

  • This is a full-length, CBSE board-hard test covering the whole chapter. All questions are compulsory.
  • Section A carries 1 mark each (including an assertion-reason item), Section B 2 marks, Section C 3 marks and Section D 5 marks (long answer / case based).
  • Show all working for Sections B, C and D. Final answers are given in the Answer Key.
Section A — Multiple Choice Questions 5 × 1 = 5 marks
1.
The value of $\sin^{-1}\left(\sin\dfrac{5\pi}{6}\right)+\cos^{-1}\left(\cos\dfrac{7\pi}{6}\right)$ is:
  • A.$\pi$
  • B.$\dfrac{\pi}{2}$
  • C.$\dfrac{4\pi}{3}$
  • D.$\dfrac{2\pi}{3}$
2.
The principal value of $\tan^{-1}\sqrt3-\cot^{-1}(-\sqrt3)$ is:
  • A.$\pi$
  • B.$-\dfrac{\pi}{2}$
  • C.$0$
  • D.$2\sqrt3$
3.
Assertion (A): $\sin^{-1}(\sin 2)=2$.
Reason (R): The range of the principal value branch of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$.
  • A.Both A and R are true and R is the correct explanation of A
  • B.Both A and R are true but R is not the correct explanation of A
  • C.A is true but R is false
  • D.A is false but R is true
4.
If $\sin^{-1}x+\sin^{-1}y=\dfrac{2\pi}{3}$, then $\cos^{-1}x+\cos^{-1}y$ equals:
  • A.$\dfrac{2\pi}{3}$
  • B.$\dfrac{\pi}{3}$
  • C.$\dfrac{\pi}{6}$
  • D.$\pi$
5.
The value of $\tan\left(\dfrac12\cos^{-1}\dfrac{\sqrt5}{3}\right)$ is:
  • A.$\dfrac{3+\sqrt5}{2}$
  • B.$\dfrac{3-\sqrt5}{2}$
  • C.$\dfrac23$
  • D.$\dfrac{\sqrt5-1}{2}$
Section B — Short Answer I 4 × 2 = 8 marks
6.
Find the principal value of $\cos^{-1}\left(\cos\dfrac{13\pi}{6}\right)$.
7.
Write $\tan^{-1}\left(\dfrac{\cos x}{1-\sin x}\right)$, $-\dfrac{3\pi}{2}
8.
Prove that $2\tan^{-1}\dfrac12=\tan^{-1}\dfrac43$.
9.
Evaluate $\sin\left[\dfrac{\pi}{3}-\sin^{-1}\left(-\dfrac12\right)\right]$.
Section C — Short Answer II 4 × 3 = 12 marks
10.
Prove that $\tan^{-1}\dfrac14+\tan^{-1}\dfrac29=\dfrac12\cos^{-1}\dfrac35$.
11.
Solve for $x$: $\tan^{-1}\dfrac{x-1}{x-2}+\tan^{-1}\dfrac{x+1}{x+2}=\dfrac{\pi}{4}$.
12.
Simplify $\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)$, $x\ne0$.
13.
Show that $\sin^{-1}\dfrac35-\sin^{-1}\dfrac{8}{17}=\cos^{-1}\dfrac{84}{85}$.
Section D — Long Answer / Case Based 2 × 5 = 10 marks
14.
Solve: $2\tan^{-1}(\cos x)=\tan^{-1}(2\operatorname{cosec}x)$ for $0
15.
Case study. A picture $3$ m tall is hung on a wall so that its lower edge is $1$ m above the eye level of an observer standing $x$ m ($x>0$) from the wall. Let $\theta$ be the angle the picture subtends at the observer's eye.
(a) Show that $\theta=\tan^{-1}\dfrac{3x}{x^2+4}$. [2]
(b) Find $\theta$ when $x=2$, in the form $\sin^{-1}(\cdot)$. [1]
(c) Using $x^2+4\ge4x$, show that $\theta\le\tan^{-1}\dfrac34$ for every $x>0$, and state where the observer should stand for the best view. [2]

Answer Key

Section A — Multiple Choice Questions
  1. (A) $\pi$
  2. (B) $-\dfrac{\pi}{2}$
  3. (D) A is false but R is true
  4. (B) $\dfrac{\pi}{3}$
  5. (B) $\dfrac{3-\sqrt5}{2}$
Section B — Short Answer I
  1. $\cos\dfrac{13\pi}{6}=\cos\left(2\pi+\dfrac{\pi}{6}\right)=\cos\dfrac{\pi}{6}$, and $\dfrac{\pi}{6}\in[0,\pi]$. So the principal value is $\dfrac{\pi}{6}$.
  2. $\dfrac{\cos x}{1-\sin x}=\dfrac{\cos^2\frac x2-\sin^2\frac x2}{\left(\cos\frac x2-\sin\frac x2\right)^2}=\dfrac{\cos\frac x2+\sin\frac x2}{\cos\frac x2-\sin\frac x2}=\tan\left(\dfrac{\pi}{4}+\dfrac x2\right)$. For the given range, $\dfrac{\pi}{4}+\dfrac x2\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, so the expression equals $\dfrac{\pi}{4}+\dfrac x2$.
  3. For $\theta=\tan^{-1}\frac12\in\left(0,\frac{\pi}{4}\right)$, $\tan2\theta=\dfrac{2\cdot\frac12}{1-\frac14}=\dfrac43$ and $2\theta\in\left(0,\frac{\pi}{2}\right)$, so $2\theta=\tan^{-1}\dfrac43$.
  4. $\sin^{-1}\left(-\frac12\right)=-\frac{\pi}{6}$, so the expression is $\sin\left(\frac{\pi}{3}+\frac{\pi}{6}\right)=\sin\frac{\pi}{2}=1$.
Section C — Short Answer II
  1. LHS $=\tan^{-1}\dfrac{\frac14+\frac29}{1-\frac1{18}}=\tan^{-1}\dfrac{17/36}{17/18}=\tan^{-1}\dfrac12$. For RHS let $\theta=\cos^{-1}\frac35$; then $\tan\frac{\theta}{2}=\sqrt{\dfrac{1-\frac35}{1+\frac35}}=\dfrac12$, so $\frac{\theta}{2}=\tan^{-1}\frac12$. Hence LHS = RHS.
  2. $\tan^{-1}A+\tan^{-1}B=\tan^{-1}\dfrac{A+B}{1-AB}$ gives $\dfrac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x-1)(x+1)}=1\Rightarrow\dfrac{2x^2-4}{-3}=1\Rightarrow x^2=\dfrac12$. So $x=\pm\dfrac{1}{\sqrt2}$ (for these values $AB=\frac17<1$, so the formula applies).
  3. Put $x=\tan\theta$, $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, $\theta\ne0$. Then $\dfrac{\sec\theta-1}{\tan\theta}=\dfrac{1-\cos\theta}{\sin\theta}=\tan\dfrac{\theta}{2}$, and $\frac{\theta}{2}\in\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$. So the expression is $\dfrac{\theta}{2}=\dfrac12\tan^{-1}x$.
  4. Let $\alpha=\sin^{-1}\frac35$, $\beta=\sin^{-1}\frac8{17}$; $\cos\alpha=\frac45$, $\cos\beta=\frac{15}{17}$. Then $\cos(\alpha-\beta)=\dfrac45\cdot\dfrac{15}{17}+\dfrac35\cdot\dfrac8{17}=\dfrac{60+24}{85}=\dfrac{84}{85}$. Since $\alpha>\beta$ and both lie in $\left(0,\frac{\pi}{2}\right)$, $\alpha-\beta\in\left(0,\frac{\pi}{2}\right)\subset[0,\pi]$, so $\alpha-\beta=\cos^{-1}\dfrac{84}{85}$.
Section D — Long Answer / Case Based
  1. Since $|\cos x|<1$, $2\tan^{-1}(\cos x)=\tan^{-1}\dfrac{2\cos x}{1-\cos^2x}=\tan^{-1}\dfrac{2\cos x}{\sin^2x}$. Equating with $\tan^{-1}\dfrac{2}{\sin x}$: $\dfrac{2\cos x}{\sin^2x}=\dfrac{2}{\sin x}\Rightarrow\cos x=\sin x\Rightarrow\tan x=1$. In $(0,\pi)$ this gives $x=\dfrac{\pi}{4}$. Check: both sides equal $\tan^{-1}(2\sqrt2)$.
  2. (a) The upper and lower edges are $4$ m and $1$ m above eye level, so $\theta=\tan^{-1}\dfrac4x-\tan^{-1}\dfrac1x=\tan^{-1}\dfrac{\frac4x-\frac1x}{1+\frac4{x^2}}=\tan^{-1}\dfrac{3x}{x^2+4}$. (b) At $x=2$, $\tan\theta=\dfrac68=\dfrac34$, so $\sin\theta=\dfrac35$ and $\theta=\sin^{-1}\dfrac35$. (c) Since $x^2+4\ge4x$, $\dfrac{3x}{x^2+4}\le\dfrac34$, and $\tan^{-1}$ is increasing, so $\theta\le\tan^{-1}\dfrac34$, with equality iff $x^2-4x+4=0$, i.e. $x=2$. The best view is from $2$ m away from the wall.
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