Inverse Trigonometric Functions MCQ Quiz - Objective Question with Answer for Inverse Trigonometric Functions - Download Free PDF

Concept:

tan-1 x + cot-1 x = \(\rm \frac {\pi}{2}\)

Calculation:

As we know tan-1 x + cot-1 x = \(\rm \frac {\pi}{2}\)

∴ cot (tan-1 x + cot-1 x) = cot \(\rm \frac {\pi}{2}\)= 0

Concept:

The maxima and minima of any function f(x) can be calculated by keeping it's first derivative as zero.

Now to find out whether it is a maxima or minima, we take it's double derivative now if the value is negative then it is a maxima and if it is positive it is a minima. For value equals to zero, it is unstable.

Solution: Given function,

\(f(x)=1+2\sin x+3\cos ^2x, \left(0\le x\le \frac{2x}{3}\right) \)

\(f^{\prime}(x)=2\cos x-6\sin x cos x\\ f^{\prime}(x)=0 \\ 2\cos x-6\sin x cos x=0 \\ 2\cos x(1-3\sin x)=0 \\ \implies \text{either} \cos x=0 \ \text{or} \ (1-3\sin x)=0 \\ x=\sin ^{-1}(\frac{1}{3}),\frac{\pi}{2}\)

To find out whether it is a maxima or minima, we take the double derivative.

\(f^{\prime \prime}(x)=-2\sin x-6 \cos 2x\\ \text{for value} \frac{\pi}{2},\\ f^{\prime \prime}(x)=-2\sin ( \frac{\pi}{2})-6 \cos (2 \cdot \frac{\pi}{2})\\ f^{\prime \prime}(x)=-2\sin ( \frac{\pi}{2})-6 \cos \pi\\ f^{\prime \prime}(x)=-2+6 =4\)

Hence, 4>0. It is minima at x = 90°.

The correct answer is option 1.

Explanation:

If sinθ = x ⇒ θ = sin^-1x, for θ ∈ [-π/2, π/2]

sin (sin^-1 x) =x for -π/2 ≤ x ≤ π/2

We have, \(\sin^{−1}\left(\frac{−\sqrt{3}}{2}\right)\)

= sin^-1(sin( \(\frac{-\pi}{3}\)))  ----- Since \(\frac{-\pi}{3}\) ∈ [\(\frac{-\pi}{2}\), \(\frac{\pi}{2}\)]

∴ sin-1(sin( \(\frac{-\sqrt{3}}{2}\))) = \(\frac{-\pi}{3}\)

Additional InformationPrincipal Values of Inverse Trigonometric Functions:

Calculation:

Let, cot-1\(\frac{1}{2}\) = ϕ 

⇒ cot ϕ = \(\frac{1}{2}\)

⇒ sin ϕ = \(\frac{1}{\sqrt{1+\cot^2\phi}}\) = \(\frac{2}{\sqrt{5}}\)

Let cos-1x = θ 

⇒ sec θ = \(\frac{1}{x}\)

⇒ tan θ = \(\sqrt{\sec^2\theta-1}\)

⇒ tan θ = \(\sqrt{\frac{1}{x^2}-1}\)

⇒ tan θ = \(\frac{\sqrt{1-x^2}}{x}\)

Now, (tan(cos-1x) = sin(cot-1\(\frac{1}{2}\))

⇒ (tan(tan-1\(\frac{\sqrt{1-x^2}}{x}\)) = sin(sin-1\(\frac{2}{\sqrt{5}}\))

⇒ \(\frac{\sqrt{1-x^2}}{x}\) = \(\frac{2}{\sqrt{5}}\)

⇒ \(\sqrt{(1-x^2)5}\) = 2x

Squaring both sides, we get:

(1 - x²)5 = 4x2

⇒ 9x² = 5

⇒ x = \(\frac{\sqrt{5}}{3}\) (∵ x > 0)

∴ The value of x is \(\frac{\sqrt{5}}{3}\).

The correct answer is Option 2.

Explanation:

\(\rm \cos^{-1}(\cos \left(\frac{5π}{4}\right))\)

= \(\rm \cos^{-1}(\cos \left(π+\frac{π}{4}\right))\)

= \(\rm \cos^{-1}(-\cos \left(\frac{π}{4}\right))\) (as cos(π + x) = -cos x)

= \(\rm \cos^{-1}(\cos \left(π-\frac{π}{4}\right))\) (as cos(π - x) = -cos x)

= \(\rm \cos^{-1}(\cos \left(\frac{3π}{4}\right))\)

= \(\frac{3π}{4}\) (cos^-1(cos x) = x if 0 ≤ x ≤ π)

Option (2) is true. 

Concept:

\(\rm \cot \theta = \tan \left(\dfrac{\pi}{2}-\theta\right)\)​.

\(\tan^{-1} x = \dfrac{\pi}{2}-\cot^{-1} x\)

Calculation:

4 tan-1 x + cot‑1 x = π

\( 4\tan^{-1}x+\left(\dfrac{\pi}{2}-\tan^{-1}x\right)=\pi\)

\(3\tan^{-1}x=\pi-\dfrac{\pi}{2}=\dfrac{\pi}{2}\)

\( \tan^{-1}x=\dfrac{\pi}{6}\)

\(x=\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt3}\).

Concept:

• \(\rm \tan^{-1}x + \tan^{-1}y=\tan^{-1}{x+y\over 1-xy}\)
• \(\rm cot^{-1}{x} = {\pi\over2}- \tan^{-1}{x}\)
• \(\rm 2tan^{-1}\ x =tan^{-1} ({\frac {2x}{1\ -\ x^2}})\)

Calculation:

S = \(\rm cot^{-1}{1\over3} - 2 \tan^{-1}{2\over3} \)

S = \(\rm \left[{\pi\over2}-\tan^{-1}{1\over3}\right] - \tan^{-1}{{2\times \frac{2}{3}}\over1-{(\frac{2}{3})^2}}\)

S = \(\rm \left[{\pi\over2}-\tan^{-1}{1\over3}\right] - \tan^{-1}{{\frac{4}{3}}\over1-{\frac{4}{9}}}\)

S = \(\rm {\pi\over2}-\tan^{-1}{1\over3} - \tan^{-1}{\frac{12}{5}}\)

S = \(\rm {\pi\over2}- \left[ \tan^{-1}{12\over5}+\tan^{-1}{1\over3}\right]\)

S = \(\rm {\pi\over2}- \left[ \tan^{-1}{{12\over5}+{1\over3}\over1-{12\over5}\times{1\over3}}\right]\)

S = \(\rm {\pi\over2}- \left[ \tan^{-1}{41\over3}\right]\)

S = \(\rm \cot^{-1}{41\over3}\)

Concept:

• The domain of a function f(x) is the set of values of x for which the function is defined.
• The value of sin θ always lies in the interval [-1, 1].
• sin^-1 (sin θ) = θ.
• sin (sin^-1 x) = x.

Calculation:

Let's say that sin-1 4x = θ.

⇒ sin (sin-1 4x) = sin θ

⇒ sin θ = 4x

Since, -1 ≤ sin θ ≤ 1

⇒ -1 ≤ 4x ≤ 1

⇒ \(\rm -\dfrac{1}{4}\leq x \leq \dfrac{1}{4}\)

⇒ x ∈ \(\rm \left[-\dfrac{1}{4},\dfrac{1}{4}\right]\)

∴ The domain of the function is the closed interval \(\rm \left[-\dfrac{1}{4},\dfrac{1}{4}\right]\).

Concept:

sin^-1 x + cos^-1 x = \(\rm \frac{π}{2}\)  

Calculation:

sin^-1 x + sin^-1 y = \(\rm \frac{3π}{4}\) 

⇒ \(\rm \left ( \frac{π}{2} -cos^{-1}x\right ) + \left ( \frac{π}{2}- cos^{-1}y\right ) = \frac{3π}{4}\) 

⇒ π - ( cos^-1 x + cos^-1 y ) = \(\rm \frac{3π}{4}\)

⇒ cos^-1 x + cos^-1 y = \(\rm\frac{\pi}{4}\) 

The correct option is 2.

Concept:

\({\sin ^{ - 1}}\ (sin x) = x\) if \(x \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]\) but if \(x{ \notin }\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]\), then use \(\sin x = \sin \left( {\pi - x} \right)\) to bring the value of x inside the principle branch.

Solution:

\({\sin ^{ - 1}}\left( {\sin \frac{{4\pi }}{5}} \right)\) but \(\frac{{4\pi }}{5}\notin{ }\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]\)

So, use the relation,

\(\sin \frac{{4\pi }}{5} = \sin \left( {\pi - \frac{{4\pi }}{5}} \right)\)

\(= \sin \left( {\frac{\pi }{5}} \right)\)

So,

\({\sin ^{ - 1}}\left( {\sin \frac{{4\pi }}{5}} \right) = {\sin ^{ - 1}}\left( {\sin \frac{\pi }{5}} \right)\)

\( = \frac{\pi }{5}\)

Concept:

sin-1 x + cos-1 x = π/2, x ∈ [-1, 1]

Calculation:

Given:  3 sin-1 x + cos-1 x = π 

 ⇒ 3 sin-1 x + cos-1 x = 2 sin-1 x + [sin-1 x + cos-1 x] = π 

As we know that, sin-1 x + cos-1 x = π/2, x ∈ [-1, 1]

⇒  2 sin-1 x + [π /2] = π

⇒ 2 sin-1 x = π - π/2

⇒ 2 sin-1 x = π/2

⇒ sin^-1 x = π/4

⇒ x = sin π/4 = 1/√2

Concept:

The principal solutions of a trigonometric equation are those solutions that lie between 0 and 2π.

Formula:

General solution of tan(x) = tan(α) is  given as;

x = nπ + α where α ∈ (-π/2 , π/2) and n ∈ Z.

Calculation:

Given, \(\tan x=-\frac{1}{\sqrt{3}}\)

⇒ tan(x) = tan(-π/6)

∴ α = -π/6

⇒ x = nπ + (-π/6) , n ∈ Z

Putting n = 1 and 2, we get - 

x = 5π/6 and 11π/6

Concept:

tan-1 x + cot-1 x = \(\rm \frac {π}{2}\)

Calculation:

To Find: Value of cos (2tan-1 x + 2cot-1 x)

cos (2tan-1 x + 2cot-1 x) = cos 2(tan-1 x + cot-1 x)

As we know, tan-1 x + cot-1 x = \(\rm \frac {π}{2}\)

cos (2tan-1 x + 2cot-1 x) = cos [2 × \(\rm \frac {π}{2}\) ]

= cos π 

= -1

Given:

AB = 20 cm

BC = 21 cm 

AC = 29 cm

Concept used:

Pythagoras' theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.

Calculation:

Using pythagoras theorem,

AC² = AB² + BC²

⇒ 29² = 20² + 21²

ΔABC is a right angled triangle.

⇒ cot C = BC/AB = 21/20

⇒ cosec C = AC/AB = 29/20

⇒ tan A = BC/AB = 21/20

cot C + cosec C - 2tan A = 21/20 + 29/20 - 2 × 21/20

⇒ 8/20

⇒ 2/5

So, the value of cot C + cosec C - 2tan A = 2/5

Concept:

\(\rm tan^{-1}(x)+tan^{-1}({y})\) = \(\rm tan^{-1}\frac{x+y}{1-x\times y}\)

\(\rm cot^{-1}x=\) \(\rm tan^{-1}({1\over x})\)

Calculation:

Given, \(\rm tan^{-1}x+cot^{-1}x=\frac{\pi}{2}\)

⇒ \(\rm tan^{-1}(x)+tan^{-1}({1\over x})=\frac{\pi}{2}\)

⇒ \(\rm tan^{-1}(x)+tan^{-1}({1\over x})=\frac{\pi}{2}\)

⇒ \(\rm tan^{-1}\frac{x+\frac{1}{x}}{1-x\times \frac{1}{x}} = \frac{\pi}{2}\)

⇒ \(\rm tan^{-1}\frac{x+\frac{1}{x}}{0} = \frac{\pi}{2}\)

⇒ \(\rm tan^{-1}\frac{x+\frac{1}{x}}{0} = \frac{\pi}{2}\)

⇒ \(\rm tan^{-1}({\infty}) = \frac{\pi}{2}\)

This is true for all x ∈ R