25. Inverse Circular Functions

Definition: Inverse functions related to trigonometrical ratios are called inverse trigonometric functions. The definition of different inverse trigonometric functions is given below:

If sinθ=x,\sin\theta = x, then θ=sin1x,\theta = \sin^{-1}x, provided 1x1-1\leq x\leq 1 and π2θπ2-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}

If cosθ=x,\cos\theta = x, then θ=cos1x,\theta = \cos^{-1}x, provoded 1x1-1\leq x\leq 1 and 0θπ0\leq \theta \leq \pi

If tanθ=x,\tan\theta = x, then θ=tan1x,\theta = \tan^{-1}x, provided <x<-\infty < x < \infty and π2<θ<π2-\frac{\pi}{2} < \theta < \frac{\pi}{2}

If cotθ=x,\cot\theta = x, then θ=cot1x,\theta = \cot^{-1}x, provided <x<-\infty < x < \infty and 0<θ<π0 < \theta < \pi

If secθ=x,\sec\theta = x, then θ=sec1x,\theta = \sec^{-1}x, provided x1x \leq -1 or x1x\geq 1 and 0θπ,θπ20\leq \theta \leq \pi, \theta \neq \frac{\pi}{2}

If cosecθ=x,\cosec\theta = x, then θ=cosec1x,\theta = \cosec^{-1}x, provided x1x \leq -1 or x1x\geq 1 and x1x\leq -1 or x1x\geq 1 and π2θπ2,θ0-\frac{\pi}{2}\leq \theta\leq \frac{\pi}{2}, \theta\neq 0

Note: In the above definition restrictions on θ\theta are due to the consideration of principal values of inverse terms. If these restrictions are removed, the terms will represent inverse trigonometrical relation and not functions.

Notations: I. Arc sinx\sin x denotes the sine inverse of xx [General value]

arcsinx\arcsin x denotes the sine inverse of xx [Principal value]

II. sin1x\sin^{-1}x denotes the principal value of sine inverse xx

From the above notations three imprtant results follow:

  1. sin1x=θsinθ=x\sin^{-1}x = \theta \Rightarrow \sin \theta = x and θ\theta is the principal value.

  2. sin1x=arcsinx,cos1x=arccosx\sin^{-1}x = \arcsin x, \cos^{-1}x = \arccos x

  3. From the definition of the inverse functions, we know that if y=f(x)y = f(x) is a function then for f1f^{-1} to be a function, ff must be one-one and onto mapping.

When we consider y=Arcsinx,y = Arc\sin x, for any x[1,1]x\in[-1, 1] infinite number of values of yy are obtained and hence it does not represent inverse functions. When y=arcsinxy = \arcsin x or sin1x,\sin^{-1}x, corresponding to one value of x[1,1],x\in [-1, 1], one values of yy is obtained and hence it represents inverse trigonometric function.

Hence, for inverse trigonometric functions, consideration of principal values is essential.

25.1. Principal Value

Numerically smallest angle is known as the principal value.

Since inverse trigonometrical terms are in fact angles, definitions of principal value of inverse trigonometrical term is the same as the definition of the principal value of angles.

Suppose we have to find the principal value of sin112.\sin^{-1}\frac{1}{2}.

For this let sin112=θ\sin^{-1}\frac{1}{2} = \theta then sinθ=12\sin\theta = \frac{1}{2}

θ=,11π6,7π6,π6,5π6,\Rightarrow \theta = \ldots, -\frac{11\pi}{6}, -\frac{7\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}, \ldots

Among all these angles π6\frac{\pi}{6} is the numerically smallest angles satisfying sinθ=12\sin\theta = \frac{1}{2} and hence it is principal value.

The steps to find principal value is same as described in previous chapter.

25.2. Important Formulae

  1. sinsin1x=x,1x1\sin\sin^{-1}x = x, -1\leq x\leq 1

  2. coscos1x=x,1x1\cos\cos^{-1}x = x, -1\leq x\leq 1

  3. tantan1x=x,<x<\tan\tan^{-1}x = x, -\infty< x < \infty

  4. cotcot1x=x,<x<\cot\cot^{-1}x = x, -\infty< x < \infty

  5. secsec1x=x,x1\sec\sec^{-1}x = x, x\leq -1 or x1x\geq 1

  6. coseccosec1x=x,x1\cosec\cosec^{-1}x = x, x\leq -1 or x1x\geq 1

Proof: Let sin1x=θ\sin^{-1}x = \theta then sinθ=x\sin\theta = x

Putthing the value of θ\theta from first equation in second

sinsin1x=x\sin\sin^{-1}x = x

Other formulae can be proved in similar manner.

  1. sin1sinx=x  π2xπ2\sin^{-1}\sin x = x~\forall~-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}

  2. cos1cosx=x  0xπ\cos^{-1}\cos x = x~\forall~0\leq x\leq \pi

  3. tan1tanx=x  0<x<\tan^{-1}\tan x = x~\forall~0<x<\infty

  4. cot1cotx=x  0xπ\cot^{-1}\cot x = x~\forall~0\leq x\leq \pi

  5. sec1secx=x  0xπ,xπ2\sec^{-1}\sec x = x~\forall~0\leq x\leq\pi, x\neq\frac{\pi}{2}

  6. cosec1cosecx=x  π2xπ2,x0\cosec^{-1}\cosec x = x~\forall~-\frac{\pi}{2}\leq x\leq\frac{\pi}{2}, x\neq 0

Proof: Let sin1x=θ\sin^{-1}x = \theta then sinθ=x\sin\theta = x

Substituting the value of x,x, we get

sin1sinθ=θ\sin^{1}\sin\theta = \theta

Replacing θ\theta by x,x, we get

sin1sinx=x\sin^{-1}\sin x = x

Other formulae can be proved in similar manner.

  1. sin1x+cos1x=π2  1x1\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}~\forall~-1\leq x\leq 1

  2. tan1x+cot1x=π2 xR\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}~\forall x \in R

  3. sec1x+cosec1x=π2 x1\sec^{-1}x + \cosec^{-1}x = \frac{\pi}{2}~\forall x\leq -1 or x1x\geq 1

Proof: Let sin1x=θ\sin^{-1}x = \theta then sinθ=x\sin\theta = x

cos(π2θ)=xπ2θ=cos1x\cos\left(\frac{\pi}{2} - \theta\right) = x \Rightarrow \frac{\pi}{2} - \theta = \cos^{-1}x

cos1x+θ=π2sin1x+cos1x=π2\cos^{-1}x + \theta = \frac{\pi}{2}\Rightarrow \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}

Similarly other results can be proven.

  1. sin1x=cosec11x,1x1\sin^{-1}x = \cosec^{-1}\frac{1}{x}, -1\leq x\leq 1

    cosec1x=sin112,x1\cosec^{-1}x = \sin^{-1}\frac{1}{2}, x\leq -1 or x>1x > 1

  2. cos1x=sec11x,1x1\cos^{-1}x = sec^{-1}\frac{1}{x}, -1\leq x\leq 1

    sec1x=cos11x,x1\sec^{-1}x = \cos^{-1}\frac{1}{x}, x\leq -1 or x1x\geq 1

  3. tan1x=cot11x,x>0\tan^{-1}x = \cot^{-1}\frac{1}{x}, x>0

    cot1x=tan11x,x>0\cot^{-1}x = \tan^{-1}\frac{1}{x}, x >0

    tan1x=cot11xπ,x<0\tan^{-1}x = \cot^{-1}\frac{1}{x} - \pi, x < 0

    cot1x=π+tan11x,x<0\cot^{-1}x = \pi + \tan^{-1}\frac{1}{x}, x < 0

Proof: Let sin1x=θ\sin^{-1}x = \theta then sinθ=x\sin\theta = x

cosecθ=1x\Rightarrow \cosec\theta = \frac{1}{x}

θ=cosec11x\Rightarrow \theta = \cosec^{-1}\frac{1}{x}

sin1x=cosec11x\Rightarrow \sin^{-1}x = \cosec^{-1}\frac{1}{x}

Other results can be proven similarly.

  1. sin1x=cos11x2,  0x1\sin^{-1}x = \cos^{-1}\sqrt{1 - x^2}, ~\forall~0\leq x\leq 1

  2. sin1x=cos11x2 forall 1x<0\sin^{-1}x = -\cos^{-1}\sqrt{1 - x^2}~forall~-1\leq x< 0

Proof: Let sin1x=θ\sin^{-1}x = \theta then sinθ=x\sin\theta = x

cos2θ=1x2cosθ=±1x2\Rightarrow \cos^2\theta = 1 - x^2 \Rightarrow \cos\theta = \pm\sqrt{1 - x^2}

Principal values of sin1x\sin^{-1}x lies between π2-\frac{\pi}{2} and π2\frac{\pi}{2}

In this interval cosθ\cos\theta is +ve

sin1x=cos11x2\Rightarrow \sin^{-1}x = \cos^{-1}\sqrt{1 - x^2}

For 1x<0-1\leq x < 0 sin1x\sin^{-1}x will be negative angle while cos11x2\cos^{-1}\sqrt{1 - x^2} will be positive angle. Hence to balance that we need to used a negative sign for this.

  1. sin1(x)=sin1x\sin^{-1}(-x) = -\sin^{-1}x

  2. cos1(x)=πcos1x\cos^{-1}(-x) = \pi - \cos^{-1}x

  3. tan1(x)=tan1x\tan^{-1}(x) = -\tan^{-1}x

  4. cot1x=πcot1x\cot^{-1}x = \pi - \cot^{-1}x

Proof: Let cos1(x)=θ\cos^{-1}(-x) = \theta then cosθ=x\cos\theta = -x

cosθ=xcos(πθ)=x-\cos\theta = x \Rightarrow \cos(\pi - \theta) = x

θ=πcos1x\therefore \theta = \pi - \cos^{-1}x

Note: cos(π+θ)\cos(\pi + \theta) is also equal to cosθ-\cos\theta but this will make principal value greater than π.\pi.

Similarly other results can be proven.

  1. tan1x+tan1y=tan1x+y1xy\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x + y}{1 - xy} where x,y>0x, y > 0 and xy<1xy < 1

    tan1x+tan1y=π+tan1x+y1xy\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\frac{x + y}{1 - xy} where x,y>0x, y > 0 and xy>1xy > 1

    tan1x+tan1y=π+tan1x+y1xy\tan^{-1}x + \tan^{-1}y = -\pi + \tan^{-1}\frac{x + y}{1 - xy} where x,y,0x, y , 0 and xy>1xy > 1

  2. tan1xtan1y=tan1xy1+xy\tan^{-1}x - \tan^{-1}y = \tan^{-1}\frac{x - y}{1 + xy} where xy>1xy > 1

Proof: Let tan1x=α\tan^{-1}x = \alpha and tan1y=β\tan^{-1}y = \beta then

tanα=x\tan\alpha = x and tanβ=y\tan\beta = y

tan(α+β)=tanα+tanβ1tanαtanβ=x+y1xy\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{x + y}{1 - xy}

α+β=tan1x+y1=xy\Rightarrow \alpha +\beta = \tan^{-1}\frac{x + y}{1 = xy}

tan1x+tan1y=tan1x+y1xy\Rightarrow \tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x + y}{1 - xy}

Case I. When x,y>0x, y > 0 and xy<1,tan1x+y1xy>0xy < 1, \tan^{-1}\frac{x + y}{1 - xy} > 0

therefore tan1x+y1xy\tan^{-1}\frac{x + y}{1 - xy} will be a positive angle.

Case II. When x,y>0x, y >0 and xy>1xy > 1 tan1x+y1xy\tan^{-1}\frac{x + y}{1 - xy} will be a negative angle.

tan1x+tan1y=π+tan1x+y1xy\therefore \tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\frac{x + y}{1 - xy}

Case III. When x,y<0x, y< 0 and xy>1xy > 1 tan1x+tan1y\tan^{-1}x + \tan^{-1}y will be a negative angle and tan1x+y1xy\tan^{-1}\frac{x + y}{1 - xy} will be a positive angle.

To balance it we will need to add π-\pi

tan1x+tan1y=π+tan1x+y1xy\therefore \tan^{-1}x + \tan^{-1}y = -\pi + \tan^{-1}\frac{x + y}{1 - xy}

Similarly other result can be proven.

  1. tan1x+tan1y+tan1z=x+y+zxyz1xyyzxz\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \frac{x + y + z - xyz}{1 - xy - yz - xz}

This can be proven that previous formula.

  1. sin1x+sin1y=sin1[x1y2+y1x2]\sin^{-1}x + \sin^{-1}y = \sin^{-1}[x\sqrt{1 - y^2} + y\sqrt{1 - x^2}] if 1x,y1-1\leq x, y\leq 1 and x2+y21x^2 + y^2\leq 1 or if xy<0xy < 0 and x2+y2>1x^2 + y^2 > 1

  2. sin1xsin1y=sin1[x1y2y1x2]\sin^{-1}x - \sin^{-1}y = \sin^{-1}[x\sqrt{1 - y^2} - y\sqrt{1 - x^2}] if 1x,y1-1\leq x, y\leq 1 and x2+y21x^2 + y^2\leq 1 or if xy>0xy > 0 and x2+y2>1x^2 + y^2 > 1

Proof: Let sin1x=α\sin^{-1}x = \alpha and sin1y=β\sin^{-1}y = \beta then sinα=x,sinβ=y.\sin\alpha = x, \sin\beta = y.

Now sin(α+β)=sinαcosβ+sinβcosα\sin(\alpha + \beta) = \sin\alpha\cos\beta + \sin\beta\cos\alpha

=sinα1sin2β+sinβ1sin2α= \sin\alpha\sqrt{1 - \sin^2\beta} + \sin\beta\sqrt{1 - \sin^2\alpha}

=x1y2+y1x2= x\sqrt{1 - y^2} + y\sqrt{1 - x^2}

α+β=sin1[x1y2+y1x2]\alpha + \beta = \sin^{-1}[x\sqrt{1 - y^2} + y\sqrt{1 - x^2}]

Similarly we can prove that sin1xsin1y=sin1[x1y2y1x2]\sin^{-1}x - \sin^{-1}y = \sin^{-1}[x\sqrt{1 - y^2} - y\sqrt{1 - x^2}]

  1. 2tan1x=sin12x1+x2,2\tan^{-1}x = \sin^{-1}\frac{2x}{1 + x^2}, where x<1|x|< 1

  2. 2tan1x=cos11x21+x2,2\tan^{-1}x = \cos^{-1}\frac{1 - x^2}{1 + x^2}, where x0x\geq 0

  3. 2tan1x=tan12x1x2,2\tan^{-1}x = \tan^{-1}\frac{2x}{1 - x^2}, where x<1|x| < 1

Proof:

  1. Let tan1x=θ\tan^{-1}x = \theta then tanθ=x\tan\theta = x

    sin2θ=2tanθ1+tan2xθ=2x1+x2\sin2\theta = \frac{2\tan\theta}{1 + \tan^2x\theta} = \frac{2x}{1 + x^2}

    2θ=sin12x1+x22tan1x=sin12x1+x2\Rightarrow 2\theta = \sin^{-1}\frac{2x}{1 + x^2}\Rightarrow 2\tan^{-1}x = \sin^{-1}\frac{2x}{1 + x^2}

    Here, π2sin1π2-\frac{\pi}{2}\leq \sin^{-1} \leq \frac{\pi}{2}

    π22tan1xπ2-\frac{\pi}{2}\leq 2\tan^{-1}x\leq \frac{\pi}{2}

    π4tan1xπ4-\frac{\pi}{4}\leq \tan^{-1}x\leq \frac{\pi}{4}

    1x1x<1-1\leq x\leq 1 \Rightarrow |x| < 1

  2. cos2θ=1tan2θ1+tan2θ=1x21+x2\cos2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \frac{1 - x^2}{1 + x^2}

    2θ=cos1(1x21+x2)\Rightarrow 2\theta = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right)

    2tan1x=cos11x21+x22\tan^{-1}x = \cos^{-1}\frac{1 - x^2}{1 + x^2}

    For x0x \geq 0 both sides will be balanced.

    For x<0,2tan1xx<0, 2\tan^{-1}x will represent a negative angle where R.H.S. will always lie between 00 and π.\pi. Hence two sides cannot be equal.

  3. tan2θ=2tanθ1tan2θ=2x1x22θ=tan12x1x2\tan2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} = \frac{2x}{1 - x^2}\Rightarrow 2\theta = \tan^{-1}\frac{2x}{1 - x^2}

    2tan1x=tan12x1x22\tan^{-1}x = \tan^{-1}\frac{2x}{1 - x^2} which holds good for x<1|x|< 1

  1. 2sin1x=sin1[2x1x2]2\sin^{-1}x = \sin^{-1}[2x\sqrt{1 - x^2}] if 12x12-\frac{1}{\sqrt{2}}\leq x\leq \frac{1}{\sqrt{2}}

  2. 2cos1x=cos1(2x21)2\cos^{-1}x = \cos^{-1}(2x^2 - 1) where 0x10\leq x \leq 1

    These can be proven like sin1x+sin1y\sin^{-1}x + \sin^{-1}y

  1. 3sin1x=sin1(3x4x3)3\sin^{-1}x = \sin^{-1}(3x - 4x^3) where 12x12-\frac{1}{2}\leq x\leq \frac{1}{2}

  2. 3cos1x=cos1(4x33x)3\cos^{-1}x = \cos^{-1}(4x^3 - 3x) where 12x1\frac{1}{2}\leq x\leq 1

  3. 3tan1x=tan13xx313x23\tan^{-1}x = \tan^{-1}\frac{3x - x^3}{1 - 3x^2} where 13<x<13-\frac{1}{\sqrt{3}}< x < \frac{1}{\sqrt{3}}

    These can be proven like previous proof.

25.3. Graph of Important Inverse Trigonometric Functions

  1. y=sin1x,1x1y = \sin^{-1}x, -1\leq x\leq 1

    y = sin^{-1}x

    Frpm this graph we observer following:

    1. Domain is 1x1-1\leq x\leq 1

    2. Range is π2yπ2-\frac{\pi}{2}\leq y \leq \frac{\pi}{2}

    3. sin1x=sin1xy=sin1x\because \sin^{-1}x = -\sin^{-1}x \therefore y = \sin^{-1}x is an odd function.

    4. It is a non-periodic function

    5. It passes through origin i.e. when x=0,y=0x = 0, y = 0

  2. y=cos1x,1x1y = \cos^{-1}x, -1\leq x\leq 1

    y = cos^{-1}x

    Follwing points can be observed from the graph:

    1. Domain is 1x1-1\leq x\leq 1

    2. Range is 0xπ0\leq x\leq \pi

    3. cos1(x)=πcos1x\because \cos^{-1}(-x) = \pi - \cos^{-1}x

      y=cos1x\Rightarrow y = \cos^{-1}x is neither odd nor even.

    4. It is a non-periodic function

  3. y=tan1x,<x<y = \tan^{-1}x, -\infty< x <\infty

    y = \tan^{-1}x

    From the graph follwing points can be observed:

    1. Domain is <x<-\infty < x < \infty

    2. Range is π2<x<π2-\frac{\pi}{2} < x < \frac{\pi}{2}

    3. y=tan1xy = \tan^{-1}x is an odd function

    4. It is a non-periodic function.

    5. It passes through origin.

  4. y=cot1x,<x<y = \cot^{-1}x, -\infty< x <\infty

    y = \cot^{-1}x

    From the graph follwing points can be observed:

    1. Domain is <x<-\infty < x < \infty

    2. Rnage is 0<y<π0<y<\pi

    3. The function is neither odd nor even.

    4. It is a non-periodic function

25.4. Problems

Evaluate the following:

  1. tan1(1)\tan^{-1}(-1)

  2. cot1(1)\cot^{-1}(-1)

  3. sin1(32)\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)

Find the value of the following:

  1. sin[π3sin112]\sin\left[\frac{\pi}{3} - \sin^{-1}\frac{-1}{2}\right]

  2. sin[cos112]\sin\left[\cos^{-1}\frac{-1}{2}\right]

  3. sin[tan1(3)+cos132]\sin\left[\tan^{-1}(-\sqrt{3}) + \cos^{-1}\frac{-\sqrt{3}}{2}\right]

  4. Evaluate tan[12cos153]\tan\left[\frac{1}{2}\cos^{-1}\frac{\sqrt{5}}{3}\right]

  5. Find the angle sin1(sin2π3)\sin^{-1}\left(\sin\frac{2\pi}{3}\right)

Find the value of the following:

  1. sin132\sin^{-1}\frac{\sqrt{3}}{2}

  2. tan113\tan^{-1}\frac{-1}{\sqrt{3}}

  3. cot1(3)\cot^{-1}(-\sqrt{3})

  4. cot1cot5π4\cot^{-1}\cot\frac{5\pi}{4}

  5. tan1(tan3π4)\tan^{-1}\left(\tan\frac{3\pi}{4}\right)

  6. sin112+cos112\sin^{-1}\frac{1}{2} + \cos^{-1}\frac{1}{2}

  7. cos[tan1(34)]\cos\left[\tan^{-1}\left(\frac{3}{4}\right)\right]

  8. cos[cos1(32)+π6]\cos\left[\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) + \frac{\pi}{6}\right]

  9. Prove that 2tan113+tan117=π42\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} = \frac{\pi}{4}

  10. Prove that tan113+tan115+tan117+tan118=π4\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{8} = \frac{\pi}{4}

  11. Prove that sin145+sin1513+sin11665=π2\sin^{-1}\frac{4}{5} + \sin^{-1}\frac{5}{13} + \sin^{-1}\frac{16}{65} = \frac{\pi}{2}

  12. Prove that 4tan115tan1170+tan1199=π44\tan^{-1}\frac{1}{5} - \tan^{-1}\frac{1}{70} + \tan^{-1}\frac{1}{99} = \frac{\pi}{4}

  13. Prove that cot19+cosec1414=π4\cot^{-1}9 + \cosec^{-1}\frac{\sqrt{41}}{4} = \frac{\pi}{4}

  14. Prove that 4(cot13+cosec15)=π4(\cot^{-1}3 + \cosec^{-1}\sqrt{5}) = \pi

  15. Prove that tan1x=2tan1[cosectan1xtancot1x]\tan^{-1}x = 2\tan^{-1}[\cosec\tan^{-1}x - \tan\cot^{-1}x]

  16. Prove that 2tan1[aba+btanx2]=cos1[b+acosxa+bcosx]2\tan^{-1}\left[\sqrt{\frac{a - b}{a + b}}\tan\frac{x}{2}\right] = \cos^{-1}\left[\frac{b + a\cos x}{a + b\cos x}\right] for 0<ba,0<b\leq a, and x0.x\geq 0.

  17. Prove that tan1xy1+xy+tan1yz1+yz+tan1zx1+zx=tan1(x2y21+x2y2)+tan1(y2z21+y2z2)+tan1(z2x21+z2x2)\tan^{-1}\frac{x - y}{1 + xy} + \tan^{-1}\frac{y - z}{1 + yz} + \tan^{-1}\frac{z - x}{1 + zx} = \tan^{-1}\left(\frac{x^2 - y^2}{1 + x^2y^2}\right) + \tan^{-1}\left(\frac{y^2 - z^2}{1 + y^2z^2}\right) + \tan^{-1}\left(\frac{z^2 - x^2}{1 + z^2x^2}\right)

  18. Prove that sincot1tancos1x=x\sin\cot^{-1}\tan\cos^{-1}x = x

  19. Prove that tan1(12tan2x)+tan1(cotx)+tan1(cot3x)=0\tan^{-1}\left(\frac{1}{2}\tan 2x\right) + \tan^{-1}(\cot x) +\tan^{-1}(\cot^3x) = 0 if π4<x<π2,=π\frac{\pi}{4}< x < \frac{\pi}{2}, = \pi if 0<x<π0<x<\pi

  20. Prove that tan112+tan113=tan135+tan114=π4\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \tan^{-1}\frac{3}{5} + \tan^{-1}\frac{1}{4} = \frac{\pi}{4}

  21. Prove that tan12ab3b+tan12ba3a=π3\tan^{-1}\frac{2a - b}{\sqrt{3}b} + \tan^{-1}\frac{2b - a}{\sqrt{3}a} = \frac{\pi}{3}

  22. Prove that tan125+tan113+tan1112=π4\tan^{-1}\frac{2}{5} + \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{12} = \frac{\pi}{4}

  23. Prove that 2tan115+tan114=tan132432\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{4} = \tan^{-1}\frac{32}{43}

  24. Prove that tan11+tan12+tan13=π=2(tan11+tan112+tan113)\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3 = \pi = 2\left(\tan^{-1}1 + \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3}\right)

  25. Prove that tan1x+cot1y=tan1xy+1yx\tan^{-1}x + \cot^{-1}y = \tan^{-1}\frac{xy + 1}{y - x}

  26. Prove that tan11x+y+tan1yx2+xy+1=cot1x\tan^{-1}\frac{1}{x + y} + \tan^{-1}\frac{y}{x^2 + xy + 1} = \cot^{-1}x

  27. Prove that 2cot15+cot17+2cot18=π/42\cot^{-1}5 + \cot^{-1}7 + 2\cot^{-1}8 = \pi/4

  28. Prove that tan1ab1+ab+tan1bc1+bc+tan1ca1+ca=0\tan^{-1}\frac{a - b}{1 + ab} + \tan^{-1}\frac{b - c}{1 + bc} + \tan^{-1}\frac{c - a}{1 + ca} = 0

  29. Prove that tan1a3b31+a3b3+tan1b3c31+b3c3+tan1c3a31+c3a3=0\tan^{-1}\frac{a^3 - b^3}{1 + a^3b^3} + \tan^{-1}\frac{b^3 - c^3}{1 + b^3c^3} + \tan^{-1}\frac{c^3 - a^3}{1 + c^3a^3} = 0

  30. Prove that cot1xy+1yx+cot1yz+1zy+cot1z=tan11x\cot^{-1}\frac{xy + 1}{y - x} + \cot^{-1}\frac{yz + 1}{z - y} + \cot^{-1}z = \tan^{-1}\frac{1}{x}

  31. Prove that cos1(cosθ+cosϕ1+cosθcosϕ)=2tan1(tanθ2tanϕ2)\cos^{-1}\left(\frac{\cos\theta + \cos\phi}{1 + \cos\theta\cos\phi}\right) = 2\tan^{-1}\left(\tan\frac{\theta}{2}\tan\frac{\phi}{2}\right)

  32. Prove that sin135+sin1817=sin17785\sin^{-1}\frac{3}{5} + \sin^{-1}\frac{8}{17} = \sin^{-1}\frac{77}{85}

  33. Prove that cos135+cos11213+cos16365=π2\cos^{-1}\frac{3}{5} + \cos^{-1}\frac{12}{13} + \cos^{-1}\frac{63}{65} = \frac{\pi}{2}

  34. Prove that sin1x+sin1y=cos1(1x21y2xy)\sin^{-1}x + \sin^{-1}y = \cos^{-1}\left(\sqrt{1 - x^2}\sqrt{1 - y^2} - xy\right) where x,y[0,1]x, y \in[0, 1]

  35. Prove that 4(sin1110+cos125)=π4\left(\sin^{-1}\frac{1}{\sqrt{10}} + \cos^{-1}\frac{2}{\sqrt{5}}\right) =\pi

  36. Prove that cos(2sin1x)=12x2\cos(2\sin^{-1}x) = 1 - 2x^2

  37. Prove that 12cos1x=sin11x2=cos11+x2=tan11x21+x\frac{1}{2}\cos^{-1}x = \sin^{-1}\sqrt{\frac{1 - x}{2}} = \cos^{-1}\sqrt{\frac{1 + x}{2}} = \tan^{-1}\frac{\sqrt{1 - x^2}}{1 + x}

  38. Prove that sin1x+cos1y=tan1xy+(1x2)(1y2)y1x2x1y2\sin^{-1}x + \cos^{-1}y = \tan^{-1}\frac{xy + \sqrt{(1 - x^2)(1 - y^2)}}{y\sqrt{1 - x^2} - x\sqrt{1 - y^2}}

  39. Prove that tan1x+tan1y=12sin12(x+y)(1xy)(1+x2)(1+y2)\tan^{-1}x + \tan^{-1}y = \frac{1}{2}\sin^{-1}\frac{2(x + y)(1 - xy)}{(1 + x^2)(1 + y^2)}

  40. Prove that 2tan1(cosectan1xtancot1x)=tan1x2\tan^{-1}(\cosec\tan^{-1}x - \tan\cot^{-1}x) = \tan^{-1}x

  41. Prove that costan1sincot1x=x2+1x2+2\cos\tan^{-1}\sin\cot^{-1}x = \sqrt{\frac{x^2 + 1}{x^2 + 2}}

  42. In any ABC\triangle ABC if A=tan12A = \tan^{-1}2 and B=tan13,B = \tan^{-1}3, prove that C=π4C = \frac{\pi}{4}

  43. If cos1x+cos1y+cos1z=π\cos^{-1}x + \cos^{-1}y + \cos^{-1}z = \pi then prove that x2+y2+z2+2xyz=1x^2 + y^2 + z^2 + 2xyz = 1

  44. If cos1x2+cos1y3=θ,\cos^{-1}\frac{x}{2}+ \cos^{-1}\frac{y}{3} = \theta, prove that 9x212xycosθ+4y2=36sin2θ9x^2 - 12xy\cos\theta + 4y^2 = 36\sin^2\theta

  45. If r=x+y+zr = x + y + z then prove that tan1xryz+tan1yrxz+tan1zrxy=π\tan^{-1}\sqrt{\frac{xr}{yz}} + \tan^{-1}\sqrt{\frac{yr}{xz}} + \tan^{-1}\sqrt{\frac{zr}{xy}} = \pi

  46. If u=cot1cos2θtan1cos2θu = \cot^{-1}\sqrt{\cos2\theta} - \tan^{-1}\sqrt{\cos2\theta} then prove that sinu=tan2θ\sin u = \tan^2\theta

  47. Solve cos1x3+cos1x=π2\cos^{-1}x\sqrt{3} + \cos^{-1}x = \frac{\pi}{2}

  48. Solve sin1x+sin12x=π3\sin^{-1}x + \sin^{-1}2x = \frac{\pi}{3}

  49. If tan1x+tan1y+tan1z=π2,\tan^{-1}x + \tan^{-1}y + \tan^{-1}z= \frac{\pi}{2}, prove that xy+yz+zx=1xy + yz + zx = 1

  50. If tan1x+tan1y+tan1z=π,\tan^{-1}x + \tan^{-1}y + \tan^{-1}z= \pi, prove that x+y+z=xyzx + y + z = xyz

  51. If sin1x+sin1y=π2,\sin^{-1}x + \sin^{-1}y = \frac{\pi}{2}, prove that x1y2+y1x2=1x\sqrt{1 - y^2} + y\sqrt{1 - x^2} = 1

  52. If sin1x+sin1y+sin1z=π,\sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \pi, prove that x1x2+y1y2+z1z2=2xyzx\sqrt{1 - x^2} + y\sqrt{1 - y^2} + z\sqrt{1 - z^2} = 2xyz

  53. Establish the relationship between tan1x,tan1y,tan1z\tan^{-1}x, \tan^{-1}y, \tan^{-1}z are in A.P. and if further x,y,zx, y, z are also in A.P. then prove that x=y=z.x = y = z.

  54. Solve for x,cot1x+sin115=π4x, \cot^{-1}x + \sin^{-1}\frac{1}{\sqrt{5}} = \frac{\pi}{4}

  55. Solve tan12x+tan13x=π4\tan^{-1}2x + \tan^{-1}3x = \frac{\pi}{4}

  56. Solve tan1x+tan12x1x2=π3\tan^{-1} x + \tan^{-1}\frac{2x}{1 - x^2} = \frac{\pi}{3}

  57. Solve tan112=cot1x+tan117\tan^{-1}\frac{1}{2} = \cot^{-1}x + \tan^{-1}\frac{1}{7}

  58. Solve tan1(x1)+tan1x+tan1(x+1)=tan13x\tan^{-1}(x - 1) + \tan^{-1}x + \tan^{-1}(x + 1) = \tan^{-1}3x

  59. Solve tan1x+1x1+tan1x1x=π+tan1(7)\tan^{-1}\frac{x + 1}{x - 1} + \tan^{-1}\frac{x - 1}{x} = \pi + \tan^{-1}(-7)

  60. Solve cot1(a1)=cot1x+cot1(a2x+1)\cot^{-1}(a - 1) = \cot^{-1}x + \cot^{-1}(a^2 - x + 1)

  61. Solve sin12α1+α2+sin12β1+β2=2tan1x\sin^{-1}\frac{2\alpha}{1 + \alpha^2} + \sin^{-1}\frac{2\beta}{1 + \beta^2} = 2\tan^{-1}x

  62. Solve cos1x21x2+1+tan12xx21=2π3\cos^{-1}\frac{x^2 - 1}{x^2 + 1} + \tan^{-1}\frac{2x}{x^2 - 1} = \frac{2\pi}{3}

  63. Solve sin12a1+a2+cos11b21+b2=2tan1x\sin^{-1}\frac{2a}{1 + a^2} + \cos^{-1}\frac{1 - b^2}{1 + b^2} = 2\tan^{-1}x

  64. Solve sin1x+sin1(1x)=cos1x\sin^{-1}x + \sin^{-1}(1 - x) = \cos^{-1}x

  65. Solve tan1ax+12sec1bx=π4\tan^{-1}ax + \frac{1}{2}\sec^{-1}bx = \frac{\pi}{4}

  66. Solve tan(cos1x)=sin(tan12)\tan(\cos^{-1}x) = \sin(\tan^{-1}2)

  67. Solve tan(sec11x)=sincos115\tan\left(\sec^{-1}\frac{1}{x}\right) = \sin\cos^{-1}\frac{1}{\sqrt{5}}

  68. Find the values of xx and yy satisfying sin1x+sin1y=2π3\sin^{-1}x + \sin^{-1}y = \frac{2\pi}{3} and cos1xcos1y=π3\cos^{-1}x - \cos^{-1}y = \frac{\pi}{3}

  69. Find the angle sin1(sin(10)\sin^{-1}(\sin(10)

  70. Using principal values, express the following as a single angle 3tan112+2tan115+sin11426553\tan^{-1}\frac{1}{2} + 2\tan^{-1}\frac{1}{5} + \sin^{-1}\frac{142}{65\sqrt{5}}

  71. Find the value of cos(2cos1x+sin1x)\cos\left(2\cos^{-1}x + \sin^{-1}x\right) at x=15x = \frac{1}{5}

  72. Show that 12cos135=tan112=π412cos145\frac{1}{2}\cos^{-1}\frac{3}{5} = \tan^{-1}\frac{1}{2} = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}\frac{4}{5}

  73. Find the greater angle between 2tan1(221)2\tan^{-1}(2\sqrt{2} - 1) and 3sin113+sin1353\sin^{-1}\frac{1}{3} + \sin^{-1}\frac{3}{5}

  74. Prove that tan1(a1xyx+a1y)+tan1(a2a11+a2a1)+tan1(a3a21+a3a2)++tan1(anan11+anan1)+tan11an=tan1xy\tan^{-1}\left(\frac{a_1x - y}{x + a_1y}\right) + \tan^{-1}\left(\frac{a_2 - a_1}{1 + a_2a_1}\right) + \tan^{-1}\left(\frac{a3 - a_2}{1 + a_3a_2}\right) + \ldots + \tan^{-1}\left(\frac{a_n - a_{n - 1}}{1 + a_na_{n - 1}}\right) + \tan^{-1}\frac{1}{a_n} = \tan^{-1}\frac{x}{y}

  75. Find the sum cot12+cot18+cot118++\cot^{-1}2 + \cot^{-1}8 + \cot^{-1}18 + \ldots + to \infty

  76. Show that the function y=2tan1x+sin12x1+x2y = 2\tan^{-1}x + \sin^{-1}\frac{2x}{1 + x^2} is constant for x1.x\geq 1. Find the value of this constant.

  77. Prove the relations cot1x0=1x02x1x2x3 to \cot^{-1}x_0 = \frac{\sqrt{1 - x_0^2}}{x_1x_2x_3\ldots\text{~to~}\infty} where the successive quantities xrx_r are connected by the relation xr+1=1+xr2x_{r + 1} = \sqrt{\frac{1 + x_r}{2}}

  78. If a,ba, b are positive quantities and if a1=a+b2,b1=a1b,a2=a1+b12,b2=a2b1a_1 = \frac{a + b}{2}, b_1 = \sqrt{a_1b}, a_2 = \frac{a_1 + b_1}{2}, b_2 = \sqrt{a_2b_1} and so on then show that limnan=limnbn=b2a2cos1ab\lim_{n\to \infty}a_n = \lim_{n\to\infty}b_n = \frac{\sqrt{b^2 - a^2}}{\cos^{-1}\frac{a}{b}}

  79. Using Mathematical Induction prove that tan113+tan117++tan11n2+n+1=tan1nn+2\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} + \ldots + \tan^{-1}\frac{1}{n^2 + n + 1} = \tan^{-1}\frac{n}{n + 2}