Inverse Trigonometric Functions - Test Papers

 CBSE Test Paper 01

Chapter 2 Inverse Trigonometric Functions

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1. The period of the function f(x) = cos4x + tan3x is 

   1. π3$\frac{\pi }{3}$
   2. π$\pi$
   3. None of these
   4. π2$\frac{\pi }{2}$

2. If 3sin−1(2x1+x2)$3{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ −4cos−1(1−x21+x2)$-4{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$+2tan−1(2x1−x2)=π3$+2{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi }{3}$. Then, x=. 

   1. 13√$\frac{1}{\sqrt{3}}$
   2. 12√$\frac{1}{\sqrt{2}}$
   3. 2
   4. 1

3. The value of tan150+cot150$\tan {15}^0+\cot {15}^0$is 

   1. 4
   2. Not defined
   3. 3–√$\sqrt{3}$
   4. 23–√$2\sqrt{3}$

4. The values of x which satisfy the trigonometric equation tan−1(x−1x−2)+tan−1(x+1x+2)=π4${\tan }^{-1}\left(\frac{x-1}{x-2}\right)+{\tan }^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi }{4}$ are: 

   1. ±2$\pm 2$
   2. ±12$\pm \frac{1}{2}$
   3. ±12√$\pm \frac{1}{\sqrt{2}}$
   4. ±2–√$\pm \sqrt{2}$

5. The minimum value of sinx - cosx is 

   1. −2–√$-\sqrt{2}$
   2. -1
   3. 0
   4. 1
6. The principle value of tan-13–√$\sqrt{3}$ is ________.
7. If y = 2 tan-1x + sin-1(2x1+x2)$\left(\frac{2x}{1+x^2}\right)$ for all x, then ________ < y < ________.
8. The value of cos (sin-1x + cos-1x), |x| ≤$\le$ 1 is ________.

9. Find the principal value of sin−1(12√)${\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)$. 

10. Write the principal value of cos−1${}^{-1}$1 [cos(680)°]. 

11. Prove that tan−1x−−√=12cos−1(1−x1+x)${\tan }^{-1}\sqrt{x}=\frac{1}{2}{\cos }^{-1}\left(\frac{1-x}{1+x}\right)$. (1)

12. Find the value of the expression tan−1(tan3π4)${\tan }^{-1}\left(\tan \frac{3\pi }{4}\right)$. 

13. Solve the equation: 2tan-1(cosx) = tan-1(2cosec x). 

14. Find the value of sin−1(sin2π3)${\sin }^{-1}\left(\sin \frac{2\pi }{3}\right)$. (2)

15. Prove that tan−1(1)+tan−1(2)+tan−1(3)=π.${\tan }^{-1}(1)+{\tan }^{-1}(2)+{\tan }^{-1}(3)=\pi .$ 

16. Solve for x, tan−1x2+tan−1x3=π4,6–√>x>0.${\tan }^{-1}\frac{x}{2}+{\tan }^{-1}\frac{x}{3}=\frac{\pi }{4},\sqrt{6}>x>0.$ 

17. Find the value of the following: tan−1[2cos(2sin−112)]${\tan }^{-1}\left[2\cos \left(2{\sin }^{-1}\frac{1}{2}\right)\right]$. 

18. Show that sin−11213+cos−145+tan−16316=π${\sin }^{-1}\frac{12}{13}+{\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{63}{16}=\pi$. 

CBSE Test Paper 01
Chapter 2 Inverse Trigonometric Functions

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Solution

1. 2. π$\pi$, Explanation: f(π)=(cos4π+tan3π)$f\left(\pi \right)=\left(cos4\pi +tan3\pi \right)$ gives the same value as f (0). Therefore, the period of the function is π$\pi$.
2. 1. 13√$\frac{1}{\sqrt{3}}$, Explanation: 3sin−1(2x1+x2)−4cos−1(1−x21+x2)+2tan−1(2x1−x2)=π3$3{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)-4{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)+2{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi }{3}$
      Put x = tanθ$\theta$
      3sin−1(2tanθ1+tan2θ)−4cos−1(1−tan2θ1+tan2θ)$3{\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)-4{\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)$+2tan−1(2tanθ1−tan2θ)=π3$+2{\tan }^{-1}\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)=\frac{\pi }{3}$
      3sin−1(sin2θ)−4cos−1(cos2θ)+2tan−1(tan2θ)=π3$3{\sin }^{-1}(\sin 2\theta )-4{\cos }^{-1}(\cos 2\theta )+2{\tan }^{-1}\left(\tan 2\theta \right)=\frac{\pi }{3}$
      3.2θ−4.2θ+2.2θ=π3⇒2θ=π3$3.2\theta -4.2\theta +2.2\theta =\frac{\pi }{3}\Rightarrow 2\theta =\frac{\pi }{3}$⇒θ=π6$\Rightarrow \theta =\frac{\pi }{6}$
      ∴tan−1x=π6⇒x=tan(π6)=13√$\therefore {\tan }^{-1}x=\frac{\pi }{6}\Rightarrow x=\tan \left(\frac{\pi }{6}\right)=\frac{1}{\sqrt{3}}$
3. 1. 4, Explanation: tan150+cot150=3√−13√+1+3√+13√−1$\tan {15}^0+\cot {15}^0=\frac{\sqrt{3}-1}{\sqrt{3}+1}+\frac{\sqrt{3}+1}{\sqrt{3}-1}$
      =(3√−1)2+(3√+1)22$=\frac{{(\sqrt{3}-1)}^2+{(\sqrt{3}+1)}^2}{2}$ = 82=4$\frac{8}{2}=4$
4. 3. ±12√$\pm \frac{1}{\sqrt{2}}$, Explanation: tan−1(x−1x−2)+tan−1(x+1x+2)=π4${\tan }^{-1}\left(\frac{x-1}{x-2}\right)+{\tan }^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi }{4}$
      tan−1[(x−1x−2)+(x+1x+2)1−(x−1x−2)(x+1x+2)]=Π4${\tan }^{-1}\left[\frac{\left(\frac{x-1}{x-2}\right)+\left(\frac{x+1}{x+2}\right)}{1-\left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)}\right]=\frac{\mathrm{\Pi }}{4}$
      tan−1[(x−1)(x+2)+(x+1)(x−2)(x−2)(x+2)−(x+1)(x−1)]=Π4${\tan }^{-1}\left[\frac{\left(x-1\right)\left(x+2\right)+\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)-\left(x+1\right)\left(x-1\right)}\right]=\frac{\mathrm{\Pi }}{4}$
      (x2+x−2+x2−x−2x2−4−x2+1)=tan−1(Π4)$\left(\frac{x^2+x-2+x^2-x-2}{x^2-4-x^2+1}\right)={\tan }^{-1}\left(\frac{\mathrm{\Pi }}{4}\right)$
      (2x2−4−3)=1$\left(\frac{2x^2-4}{-3}\right)=1$
      ∴2x2−4=−3$\therefore 2x^2-4=-3$
      ⇒2x2=1$\Rightarrow 2x^2=1$
      x=±12√$x=\pm \frac{1}{\sqrt{2}}$
5. 1. −2–√$-\sqrt{2}$, Explanation: Since, range of sine function and cosine function is [-1,1]. But, sine is increasing function and cosine is decreasing function. Therefore, the lowest that both together can attain is −450$-{45}^0$.
      (−12√)+(−12√)=−2–√$\left(-\frac{1}{\sqrt{2}}\right)+\left(-\frac{1}{\sqrt{2}}\right)=-\sqrt{2}$
6. π3$\frac{\pi }{3}$
7. -2π$\pi$, 2π$\pi$
8. 0
9. Let sin−1(12√)=θ${\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)=\theta$
   ⇒sinθ=12√$\Rightarrow \sin \theta =\frac{1}{\sqrt{2}}$
   We know that θ∈[−π2,π2]$\theta \in \left[\frac{-\pi }{2},\frac{\pi }{2}\right]$
   ⇒sinθ=sinπ4$\Rightarrow \sin \theta =\sin \frac{\pi }{4}$ ⇒θ=π4$\Rightarrow \theta =\frac{\pi }{4}$
   Therefore, principal value of sin−1(12√)${\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)$ is π4$\frac{\pi }{4}$
10. We know that, principal value branch of cos−1${\cos }^{-1}$ x is [0, 180°].
    Since, 680° ∈$\in$ [0,180°], so write 680° as 2 ×$\times$ 360°-40°
    Now, cos−1${\cos }^{-1}$[cos (680)°] = cos−1${\cos }^{-1}$ [cos(2 ;×$\times$ 360°-40°)]
    = cos−1${\cos }^{-1}$(cos40°) [∵cos(4π−θ)=cosθ]$[\because \cos (4\pi -\theta )=\cos \theta ]$
    Since, 40°∈$\in$ [0,180°]
    ∴cos−1$\therefore {\cos }^{-1}$[cos(680°)] = 40°
    [∵cos−1(cosθ)=θ;∀θ∈[0,180∘]]$\left[\because {\cos }^{-1}(\cos \theta )=\theta ;\mathrm{\forall }\theta \in \left[0,{180}^{\circ }\right]\right]$
    which is the required principal value.
11. LHS = tan−1x−−√${\tan }^{-1}\sqrt{x}$
    Let tanθ=x−−√$tan\theta =\sqrt{x}$
    tan2θ=x${\tan }^2\theta =x$
    R.H.S. =12cos−1(1−tan2θ1+tan2θ)$=\frac{1}{2}{\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)$
    =12cos−1(cos2θ)=12×2θ=θ$=\frac{1}{2}{\cos }^{-1}\left(\cos 2\theta \right)=\frac{1}{2}\times 2\theta =\theta$
    =tan−1x−−√$={\tan }^{-1}\sqrt{x}$
12. tan−1(tan3π4)${\tan }^{-1}\left(\tan \frac{3\pi }{4}\right)$
    =tan−1(tan4π−π4)$={\tan }^{-1}\left(\tan \frac{4\pi -\pi }{4}\right)$
    =tan−1[tan(π−π4)]$={\tan }^{-1}\left[\tan \left(\pi -\frac{\pi }{4}\right)\right]$
    =tan−1[−tanπ4]$={\tan }^{-1}\left[-\tan \frac{\pi }{4}\right]$
    =tan−1tan(−π4)$={\tan }^{-1}\tan \left(-\frac{\pi }{4}\right)$ =−π4$=-\frac{\pi }{4}$
13. 2 tan-1(cos x) = tan-1(2 cosec x)
    ⇒tan−1(2cosx1−cos2x)=tan−1(2sinx)$\Rightarrow {\tan }^{-1}\left(\frac{2\cos x}{1-{\cos }^{2}x}\right)={\tan }^{-1}\left(\frac{2}{\sin x}\right)$
    ⇒2cosx1−cos2x=2sinx$\Rightarrow \frac{2\cos x}{1-{\cos }^{2}x}=\frac{2}{\sin x}$
    ⇒cosxsinx=1$\Rightarrow \frac{\cos x}{\sin x}=1$
    ⇒$\Rightarrow$ cot x = 1 ⇒x=π4$\Rightarrow x=\frac{\pi }{4}$
14. sin−1(sin2π3)${\sin }^{-1}\left(\sin \frac{2\pi }{3}\right)$
    =sin−1(sin3π−π3)$={\sin }^{-1}\left(\sin \frac{3\pi -\pi }{3}\right)$
    =sin−1[sin(π−π3)]$={\sin }^{-1}\left[\sin \left(\pi -\frac{\pi }{3}\right)\right]$
    =sin−1sinπ3$={\sin }^{-1}\sin \frac{\pi }{3}$ =π3$=\frac{\pi }{3}$
15. To prove, tan−1(1)+tan−1(2)+tan−1(3)=π$ta{n}^{-1}(1)+ta{n}^{-1}(2)+ta{n}^{-1}(3)=\pi$
    LHS = tan−1(1)+tan−1(2)+tan−1(3)$ta{n}^{-1}(1)+ta{n}^{-1}(2)+ta{n}^{-1}(3)$
    =tan−1(tanπ4)+π2−cot−1(2)+π2−cot−1(3)$={\tan }^{-1}\left(\tan \frac{\pi }{4}\right)+\frac{\pi }{2}-{\cot }^{-1}(2)+\frac{\pi }{2}-{\cot }^{-1}(3)$ [∵tan−1x+cot−1x=π2]$\left[\because {\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}\right]$
    =π4+π−[cot−1(2)+cot−1(3)]$=\frac{\pi }{4}+\pi -\left[{\cot }^{-1}(2)+{\cot }^{-1}(3)\right]$[∵tan−1(tanθ)=θ;∀θ∈(−π2,π2)]$\left[\because {\tan }^{-1}(\tan \theta )=\theta ;\mathrm{\forall }\theta \in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)\right]$
    =5π4−[tan−1(12)+tan−1(13)]$=\frac{5\pi }{4}-\left[{\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{1}{3}\right)\right]$[∵cot−1x=tan−11x,x>0]$\left[\because {\cot }^{-1}x={\tan }^{-1}\frac{1}{x},x>0\right]$
    =5π4−[tan−1(12+131−12⋅13)]$=\frac{5\pi }{4}-\left[{\tan }^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot \frac{1}{3}}\right)\right]$[∵tan−1x+tan−1y=tan−1(x+y1−xy), if xy<1]$\left[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right),\text{ if }xy<1\right]$
    =5π4−tan−1(5/65/6)$=\frac{5\pi }{4}-{\tan }^{-1}\left(\frac{5/6}{5/6}\right)$
    =5π4−tan−1(1)=5π4−π4=4π4=π$=\frac{5\pi }{4}-{\tan }^{-1}(1)=\frac{5\pi }{4}-\frac{\pi }{4}=\frac{4\pi }{4}=\pi$= RHS (Hence Proved)
16. Here, we have to find the value of x .Now, we are given that
    tan−1x2+tan−1x3${\tan }^{-1}\frac{x}{2}+{\tan }^{-1}\frac{x}{3}$=π4,6–√>x>0$=\frac{\pi }{4},\sqrt{6}>x>0$
    ⇒tan−1(x2+x31−x26)=π4$\Rightarrow {\tan }^{-1}\left(\frac{\frac{x}{2}+\frac{x}{3}}{1-\frac{x^2}{6}}\right)=\frac{\pi }{4}$[∵tan−1x+tan−1y=tan−1(x+y1−xy);xy<1]$\left[\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right);xy<1\right]$
    ⇒3x+2x66−x26=tanπ4$\Rightarrow \frac{\frac{3x+2x}{6}}{\frac{6-x^2}{6}}=\tan \frac{\pi }{4}$ { taking tan on both sides}
    ⇒5x6−x2=1[∵tanπ4=1]$\Rightarrow \frac{5x}{6-x^2}=1\left[\because \tan \frac{\pi }{4}=1\right]$
    ⇒$\Rightarrow$ 5x = 6-x2
    ⇒$\Rightarrow$ x2 + 5x - 6 = 0
    ⇒$\Rightarrow$ x2 + 6x - x - 6 = 0
    ⇒$\Rightarrow$ x (x + 6) - 1 (x + 6) = 0
    ⇒$\Rightarrow$ (x-1) (x + 6) = 0
    ∴$\therefore$ x = 1 or - 6
    But it is given that, 6–√ > x > 0 ⇒x > 0$\sqrt{\text{6}}\text{ > x > 0 }\Rightarrow \text{x > 0}$
    ∴$\therefore$ x = - 6 is rejected.
    Hence, x = 1 is the only solution of the given equation.
17. tan−1[2cos(2sin−112)]${\tan }^{-1}\left[2\cos \left(2{\sin }^{-1}\frac{1}{2}\right)\right]$
    =tan−1[2cos(2sin−1sinπ6)]$={\tan }^{-1}\left[2\cos \left(2{\sin }^{-1}\sin \frac{\pi }{6}\right)\right]$
    =tan−1[2cos(2×π6)]$={\tan }^{-1}\left[2\cos \left(2\times \frac{\pi }{6}\right)\right]$
    =tan−1[2cosπ3]$={\tan }^{-1}\left[2\cos \frac{\pi }{3}\right]$
    =tan−1[2×12]$={\tan }^{-1}\left[2\times \frac{1}{2}\right]$ = tan-11
    =tan−1tanπ4=π4$={\tan }^{-1}\tan \frac{\pi }{4}=\frac{\pi }{4}$
18. Let θ$\theta$ = sin-1(1213$\frac{12}{13}$)
    ⇒$\Rightarrow$ sinθ$\theta$ = 1213$\frac{12}{13}$
    ⇒1−cos2θ−−−−−−−−√=1213$\Rightarrow \sqrt{1-co{s}^2\theta }=\frac{12}{13}$
    ⇒1−cos2θ=(12)2(13)2$\Rightarrow 1-co{s}^2\theta =\frac{(12{)}^2}{(13{)}^2}$
    ⇒cos2θ=(5)2(13)2$\Rightarrow co{s}^2\theta =\frac{(5{)}^2}{(13{)}^2}$
    ⇒cosθ=513$\Rightarrow cos\theta =\frac{5}{13}$
    Since, tanθ$\theta$ = sinθcosθ=1213513=125$\frac{sin\theta }{cos\theta }=\frac{\frac{12}{13}}{\frac{5}{13}}=\frac{12}{5}$
    ⇒θ=tan−1(125)$\Rightarrow \theta =ta{n}^{-1}(\frac{12}{5})$
    Thus, θ=sin−1(1213)=tan−1(125)$\theta =si{n}^{-1}(\frac{12}{13})=ta{n}^{-1}(\frac{12}{5})$
    Similarly, cos−1(513)=tan−1(125)$co{s}^{-1}(\frac{5}{13})=ta{n}^{-1}(\frac{12}{5})$
    We have, LHS = sin−1(1213)+cos−1(45)+tan−1(6316)${\sin }^{-1}\left(\frac{12}{13}\right)+{\cos }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)$
    =tan−1(125)+tan−1(34)+tan−1(6316)$={\tan }^{-1}\left(\frac{12}{5}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)$
    =[tan−1(125)+tan−1(34)]+tan−1(6316)$=\left[{\tan }^{-1}\left(\frac{12}{5}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)\right]+{\tan }^{-1}\left(\frac{63}{16}\right)$
    {since 125×34=95>1,$\frac{12}{5}\times \frac{3}{4}=\frac{9}{5}>1,$ therefore , tan-1A + tan-1B = π+tan−1A+_B1−AB$\pi +{\tan }^{-1}\frac{A+\mathrm{\_}B}{1-AB}$)
    =π+tan−1(125+341−(125)(34))+tan−1(6316)$\pi +{\tan }^{-1}\left(\frac{\frac{12}{5}+\frac{3}{4}}{1-\left(\frac{12}{5}\right)\left(\frac{3}{4}\right)}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)$
    = π+tan−1(−6316)+tan−1(6316)$\pi +{\tan }^{-1}\left(-\frac{63}{16}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)$
    =π−tan−1(6316)+tan−1(6316)$=\pi -{\tan }^{-1}\left(\frac{63}{16}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)$ = π$\pi$ Hence Proved.