Inverse Trigonometric Functions MCQ Quiz in मल्याळम - Objective Question with Answer for Inverse Trigonometric Functions - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

If cosθ = x ⇒ θ = cos^-1x, for θ ∈ [0, π]

cos^-1 (cos x) =x for 0 ≤ x ≤ π 

Let f(x) = cos–1(2x – 1)

We know that cos^-1x is defined for x ∈ [-1, 1]

∴ f(x) is defined for 

⇒ -1 ≤ (2x - 1) ≤ 1

⇒ 0 ≤ 2x ≤ 2

⇒ 0 ≤ x ≤ 1

∴ x ∈ [0, 1]

Concept:

• ${\sin }^{-1}\mathrm{x}-{\sin }^{-1}\mathrm{y}={\sin }^{-1}\left(\mathrm{x}\sqrt{1-{\mathrm{y}}^2}-\mathrm{y}\sqrt{1-{\mathrm{x}}^2}\right)$
• ${\sin }^{-1}\mathrm{x}+{\sin }^{-1}\mathrm{y}={\sin }^{-1}\left(\mathrm{x}\sqrt{1-{\mathrm{y}}^2}+\mathrm{y}\sqrt{1-{\mathrm{x}}^2}\right)$

Calculation:

We have to find the value of  ${\sin }^{-1}\left(\frac{4}{5}\right)-{\sin }^{-1}\left(\frac{3}{5}\right)$

We know that ${\sin }^{-1}\mathrm{x}-{\sin }^{-1}\mathrm{y}={\sin }^{-1}\left(\mathrm{x}\sqrt{1-{\mathrm{y}}^2}-\mathrm{y}\sqrt{1-{\mathrm{x}}^2}\right)$

$\therefore {\sin }^{-1}\left(\frac{4}{5}\right)-{\sin }^{-1}\left(\frac{3}{5}\right)={\sin }^{-1}(\left(\frac{4}{5}\sqrt{1-{\left(\frac{3}{5}\right)}^2}\right)-\left(\frac{3}{5}\sqrt{1-{\left(\frac{4}{5}\right)}^2}\right))$

$={\sin }^{-1}(\left(\frac{4}{5}\sqrt{1-\frac{9}{25}}\right)-\left(\frac{3}{5}\sqrt{1-\frac{16}{25}}\right))$

$={\sin }^{-1}(\left(\frac{4}{5}\sqrt{\frac{25-9}{25}}\right)-\left(\frac{3}{5}\sqrt{\frac{25-16}{25}}\right))$

$={\sin }^{-1}(\left(\frac{4}{5}\sqrt{\frac{16}{25}}\right)-\left(\frac{3}{5}\sqrt{\frac{9}{25}}\right))$

$={\sin }^{-1}(\left(\frac{4}{5}\sqrt{{\left(\frac{4}{5}\right)}^2}\right)-\left(\frac{3}{5}\sqrt{{\left(\frac{3}{5}\right)}^2}\right))$

$={\sin }^{-1}((\frac{4}{5}\times \frac{4}{5})-(\frac{3}{5}\times \frac{3}{5}))$

$={\sin }^{-1}(\left(\frac{16}{25}\right)-\left(\frac{9}{25}\right))={\sin }^{-1}\left(\frac{7}{25}\right)$

Concept:

${\sin }^{-1}\mathrm{x}+{\cos }^{-1}\mathrm{x}=\frac{\pi }{2}$

${\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}=1$

Calculation:

Given ${\sin }^{-1}\mathrm{x}+{\sin }^{-1}\mathrm{y}=\frac{\pi }{2}$

${\sin }^{-1}\mathrm{x}=\frac{\pi }{2}-{\sin }^{-1}\mathrm{y}$

sin^-1 x = cos^-1 y 

Let x = sin A

sin-1 (sin A) = cos-1 y 

A = cos-1 y 

y = cos A

y = $\sqrt{1-{\sin }^2\mathrm{A}}$

y² = 1 - x²

x² = 1 - y² 

Concept:

sec^-1 x = cos^-1 (1/x)

cosec^-1 (x) + sec^-1 (x) = π / 2

Calculation:

As we know that, 

sec-1 x = cos-1 (1/x)

$\mathrm{c}\mathrm{o}{\mathrm{s}}^{-1}\left(\frac{1}{\sqrt{5}}\right)=\theta$

$\Rightarrow \mathrm{s}\mathrm{e}{\mathrm{c}}^{-1}\left(\sqrt{5}\right)=\theta$

As we know that,

cosec-1 (x) + sec-1 (x) = π / 2

$\Rightarrow \frac{\pi }{2}-\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}{\mathrm{c}}^{-1}\left(\sqrt{5}\right)=\theta$

$\therefore \mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}{\mathrm{c}}^{-1}\left(\sqrt{5}\right)=\frac{\pi }{2}-\theta$

Explanation:

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

sin^-1 (sin x) =x for -π /2 ≤ x ≤ π/2

Given that, f(x) = sin^-1$\sqrt{x-1}$

We know that sin^-1x is defined for x ∈ [-1, 1]

∴ f(x) = sin-1$\sqrt{x-1}$ is defined for 

⇒ 0 ≤ $\sqrt{x-1}$ ≤ 1

⇒ 0 ≤ x -1 ≤ 1

⇒ 1 ≤ x ≤ 2

∴ x ∈ [1, 2]

Explanation:

sin^-1y is defined for y ∈ [-1, 1]

∴ sin-1 $\left(\frac{x+1}{3}\right)$ is defined for $\frac{x+1}{3}\in [-1,1]$

⇒ $-1\le \frac{x+1}{3}\le 1$

⇒ $-3\le x+1\le 3$

⇒ $-3-1\le x\le 3-1$

⇒ $-4\le x\le 2$

Hence, the domain of sin-1 $\left(\frac{x+1}{3}\right)$ is [-4, 2].

Concept:

The domain of inverse sine function, sin x is $x\in \left[-1,1\right]$

Calculation:

The domain of the function ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$ is calculated as follows:

$1-x^2\ge 0$

⇒ $x\in \left[-1,1\right]$      ...(1)

Also, 

$-1\le 2x\sqrt{1-x^2}\le 1$

$-\frac{1}{2}\le x\sqrt{1-x^2}\le \frac{1}{2}$

Square it to get x,

$0\le x^2\left(1-x^2\right)\le \frac{1}{4}$

Now, 

$x^2\left(1-x^2\right)\ge 0$ and $x^2\left(1-x^2\right)\le \frac{1}{4}$

Here we have to find the common values of x.

For, $x^2\left(1-x^2\right)\ge 0$

Here, the values of x for which LHS will change its sign will be -1 and 1 so the values of x for the above inequality,

⇒ $x\in \left[-1,1\right]$      ....(2)

For,

$x^2\left(1-x^2\right)\le \frac{1}{4}$

$t-t^2-\frac{1}{4}\le 0$

${\left(t-\frac{1}{2}\right)}^2\le 0$

$t\le \frac{1}{2}$

$x^2\le \frac{1}{\sqrt{2}}$

$x\in \left[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right]$        ....(3)

Take, all the common intervals from equations 1, 2, and 3,

We will get, 

$\Rightarrow x\in \left[-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right]$

Concept:

Inverse trigonometric function for negative argument

Values of trigonometric function

Calculation:

Given: $\mathrm{s}\mathrm{i}\mathrm{n}(\frac{\pi }{3}-\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}(-\frac{1}{2}))$

= $\mathrm{s}\mathrm{i}\mathrm{n}(\frac{\pi }{3}+\mathrm{s}\mathrm{i}{\mathrm{n}}^{-1}(\frac{1}{2}))$                            (∵ sin^-1 (- x) = - sin^-1 x)

= $\mathrm{s}\mathrm{i}\mathrm{n}(\frac{\pi }{3}+\frac{\pi }{6})$

= $\mathrm{s}\mathrm{i}\mathrm{n}\frac{\pi }{2}$

= 1

Given:

2sin(3x -15)° = 1

Calculation:

sin(3x - 15)° = 1/2

⇒ sin (3x - 15)° = sin 30° 

⇒ (3x - 15)° = 30° 

⇒ 3x = 30° + 15° = 45° 

⇒ x = $\frac{45}{3}$ = 15° 

Then,

cos²(2x +15)° + cot2 (x +15)°

cos2(2 × 15 +15)° + cot2 (15 +15)°

⇒ cos²(45)° + cot²(30)°

⇒ $(\frac{1}{\sqrt{2}}{)}^2$ + $(\sqrt{3}{)}^2$

⇒ $\frac{1}{2}$ + 3 = $\frac{7}{2}$ 

∴ cos2(2x +15)° + cot2 (x +15)° = $\frac{7}{2}$

Concept:

Calculation: 

Using the concepts above, we can find the principal value of the given expression as follows:

${\sin }^{-1}(-{\displaystyle \frac{\sqrt{3}}{2}})+{\cos }^{-1}(\cos \left({\displaystyle \frac{7\pi }{6}}\right))$

$=-{\sin }^{-1}{\displaystyle \frac{\sqrt{3}}{2}}+{\cos }^{-1}(\cos (\pi +{\displaystyle \frac{\pi }{6}}))$

$=-{\sin }^{-1}{\displaystyle \frac{\sqrt{3}}{2}}+{\cos }^{-1}(-\cos {\displaystyle \frac{\pi }{6}})$

$=-{\displaystyle \frac{\pi }{3}}+(\pi -{\displaystyle \frac{\pi }{6}})$

$={\displaystyle \frac{\pi }{2}}$.