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

Concept:

${\tan }^{-1}\mathrm{x}+{\tan }^{-1}\mathrm{y}={\tan }^{-1}\left[\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{x}\mathrm{y}}\right]$

$\frac{\mathrm{d}({\tan }^{-1}\mathrm{x})}{\mathrm{d}\mathrm{x}}=\frac{1}{1+{\mathrm{x}}^2}$

Calculation:

Given: y = ${\tan }^{-1}\left[\frac{8\mathrm{x}}{1-15{\mathrm{x}}^2}\right]$

$\mathrm{y}={\tan }^{-1}\left[\frac{5\mathrm{x}+3\mathrm{x}}{1-5\mathrm{x}\cdot 3\mathrm{x}}\right]$

As we know that, ${\tan }^{-1}\mathrm{x}+{\tan }^{-1}\mathrm{y}={\tan }^{-1}\left[\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{x}\mathrm{y}}\right]$

So, $\mathrm{y}={\tan }^{-1}\left[\frac{5\mathrm{x}+3\mathrm{x}}{1-5\mathrm{x}\cdot 3\mathrm{x}}\right]$= tan-1 5x + tan-1 3x

Differentiating with respect to x, we get

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}({\tan }^{-1}5\mathrm{x})}{\mathrm{d}\mathrm{x}}+\frac{\mathrm{d}({\tan }^{-1}3\mathrm{x})}{\mathrm{d}\mathrm{x}}$

$=\frac{5}{1+(5\mathrm{x}{)}^2}+\frac{3}{1+(3\mathrm{x}{)}^2}$

$=\frac{5}{1+25{\mathrm{x}}^2}+\frac{3}{1+9{\mathrm{x}}^2}$

$y={\sin }^2({\cot }^{-1}\sqrt{{\displaystyle \frac{1+x}{1-x}}})$

Let, $\cot \theta =\sqrt{{\displaystyle \frac{1+x}{1-x}}}$

${\cot }^2\theta ={\displaystyle \frac{1+x}{1-x}}$

$\Rightarrow x={\displaystyle \frac{{\cot }^2\theta -1}{{\cot }^2\theta +1}}={\displaystyle \frac{{\cos }^2\theta -{\sin }^2\theta }{{\cos }^2\theta +{\sin }^2\theta }}=\cos 2\theta$

Now,

$y={\sin }^2\theta$

$\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{dy}{d(\theta )}}\times {\displaystyle \frac{d(\theta )}{dx}}$

$\Rightarrow {\displaystyle \frac{dy}{dx}}=2\sin \theta \cos \theta \times {\displaystyle \frac{-1}{2\sin (2\theta )}}$

$\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{-1}{2}}$

Concept:

Inverse Tangent and Cotangent Relationships:

• The inverse cotangent function can be written in terms of the inverse tangent function: cot^-1 θ = (π/2) - tan^-1 θ.
• This relationship is useful in simplifying equations involving both tan^-1 and cot^-1.

Calculation:

Given the equation:

tan^-1 (2 / (3x + 1)) = cot^-1 (3 / (3x + 1))

We use the identity cot^-1 θ = (π/2) - tan^-1 θ to rewrite the equation as:

tan^-1 (2 / (3x + 1)) = (π/2) - tan^-1 (3 / (3x + 1))

Taking the tangent of both sides:

(2 / (3x + 1)) = (3 / (3x + 1))

This leads to the contradictory equation:

2 = 3

Therefore, there is no solution to this equation.

Conclusion:

The correct answer is:

• Option (1): There is no real value of x satisfying the above equation.

Calculation

Given: $y=\sqrt{{\sin }^{-1}x+y}$

Squaring both sides:

⇒ $y^2={\sin }^{-1}x+y$

Differentiating both sides with respect to x:

⇒ $2y\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}+\frac{dy}{dx}$

⇒ $2y\frac{dy}{dx}-\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$

⇒ $\frac{dy}{dx}(2y-1)=\frac{1}{\sqrt{1-x^2}}$

⇒ $\frac{dy}{dx}=\frac{1}{(2y-1)\sqrt{1-x^2}}$ 

Hence option 1 is correct.

Calculation

$y=\sqrt{\sin x+y}$

⇒ $y^2=\sin x+y$

⇒ $2y\frac{dy}{dx}=\cos x+\frac{dy}{dx}$

⇒ $2y\frac{dy}{dx}-\frac{dy}{dx}=\cos x$

⇒ $\frac{dy}{dx}(2y-1)=\cos x$

⇒ $\frac{dy}{dx}=\frac{\cos x}{2y-1}$

Substitute x = 0 and y = 1:

⇒ $\frac{dy}{dx}=\frac{\cos (0)}{2(1)-1}$

⇒ $\frac{dy}{dx}=\frac{1}{2-1}$

⇒ $\frac{dy}{dx}=\frac{1}{1}$

⇒ $\frac{dy}{dx}=1$

∴ $\frac{dy}{dx}=1$ at x = 0, y = 1

Hence option 2 is correct

Concept:

${\cot }^{-1}\mathrm{x}={\tan }^{-1}\left(\frac{1}{\mathrm{x}}\right)$

tan (tan^-1 x) = x

Calculation:

Given:

y = tan (cot⁻¹ x)                                      

 $\Rightarrow \mathrm{y}=\tan \left[{\tan }^{-1}\left(\frac{1}{\mathrm{x}}\right)\right]$               

⇒ y = 1/x                                (∵tan (tan^-1 x) = x)

Differentiating with respect to x, we get

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

At x = 1

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=-\frac{1}{1}=-1$

Concept:

• ${\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)$      ----(1)
• da/dx = 0, where a is any constant
• $\frac{d}{dx}{\tan }^{-1}x=\frac{1}{1+x^2}$      ----(2)

• $\frac{d}{dx}(a+b)=\frac{da}{dx}+\frac{db}{dx}$

Calculation:

$\frac{d}{dx}{\tan }^{-1}\left[\frac{5+x}{1-5x}\right]$

$\Rightarrow \frac{d}{dx}({\tan }^{-1}5+{\tan }^{-1}x)$      [Using (1)]

$\Rightarrow \frac{d}{dx}({\tan }^{-1}5)+\frac{d}{dx}({\tan }^{-1}x)$

$\Rightarrow 0+\frac{1}{1+x^2}$      [Using (2)]

$\Rightarrow \frac{1}{1+x^2}$

Given:

? = $\frac{\sin {33}^{\circ }\cos {57}^{\circ }+\sec {62}^{\circ }\sin {28}^{\circ }+\cos {33}^{\circ }\sin {57}^{\circ }+\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{c}{62}^{\circ }\cos {28}^{\circ }}{\tan {15}^{\circ }\tan {35}^{\circ }\tan {60}^{\circ }\tan {55}^{\circ }\tan {75}^{\circ }}$

Formula:

sin (90 - θ) = cos θ

tan (90 - θ) = cot θ

Calculation:

⇒ sin 33°. cos57° = sin(90° - 57°).cos57° = cos57°.cos57° = cos²57°

⇒ sec 62°.sin28° = 1/cos 62° × sin 28° = sin (90° - 62°) × 1/cos 62°

= cos62° × 1/cos 62° = 1

⇒ cos33°. sin 57° = sin (90° - 33°) . sin 57° = sin²57°

⇒ cosec 62°. cos 28° = cos 28° × 1/sin 62°

= cos 28° × 1/sin(90° - 28°)

= cos 28°/cos 28° = 1

⇒ tan 15°.tan 35°.tan 60°.tan 55°.tan 75° = (sin15°/cos15°) × (sin75°/cos75°) × (sin55°/cos55° ) × (sin35°/cos35°) × √3

= √3

Then,

⇒ ? = (cos²57° + 1 + sin²57° + 1)/√3

⇒ ? = 3/√3

⇒ ? = √3

∴ $\frac{\sin {33}^{\circ }\cos {57}^{\circ }+\sec {62}^{\circ }\sin {28}^{\circ }+\cos {33}^{\circ }\sin {57}^{\circ }+\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{c}{62}^{\circ }\cos {28}^{\circ }}{\tan {15}^{\circ }\tan {35}^{\circ }\tan {60}^{\circ }\tan {55}^{\circ }\tan {75}^{\circ }}$ = √3

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

• Chain Rule: $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left[\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)\right]={\mathrm{f}}^{\prime }\left(\mathrm{g}\left(\mathrm{x}\right)\right){\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$
• Product Rule: $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left[\mathrm{f}\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)\right]={\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\mathrm{g}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right){\mathrm{g}}^{\prime }\left(\mathrm{x}\right)$

Calculation:

We have to find the value of $\frac{\mathrm{d}(\sin \mathrm{x}\mathrm{t})}{\mathrm{d}\mathrm{t}}$

$$\begin{gathered} \frac{\mathrm{d}(\sin \mathrm{x}\mathrm{t})}{\mathrm{d}\mathrm{t}}=\frac{\mathrm{d}(\sin \mathrm{x}\mathrm{t})}{\mathrm{d}(\mathrm{x}\mathrm{t})}\times \frac{\mathrm{d}(\mathrm{x}\mathrm{t})}{\mathrm{d}\mathrm{t}} \\ =\cos \mathrm{x}\mathrm{t}\times \mathrm{x} \\ =\mathrm{x}\cos \mathrm{x}\mathrm{t} \end{gathered}$$

Concept:

Derivatives of Trigonometric Functions:

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\sin \mathrm{x}=\cos \mathrm{x}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\cos \mathrm{x}=-\sin \mathrm{x} \\ \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\tan \mathrm{x}={\sec }^2\mathrm{x}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\cot \mathrm{x}=-{\csc }^2\mathrm{x} \\ \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\sec \mathrm{x}=\tan \mathrm{x}\sec \mathrm{x}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\csc \mathrm{x}=-\cot \mathrm{x}\csc \mathrm{x} \end{gathered}$$

Trigonometric Formulae:

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

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

$\tan 2\mathrm{x}=\frac{2\tan \mathrm{x}}{1-{\tan }^2\mathrm{x}}$

Chain Rule of Derivatives:

• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\mathrm{f}(\mathrm{g}(\mathrm{x}))=\frac{\mathrm{d}}{\mathrm{d}\mathrm{g}(\mathrm{x})}\mathrm{f}(\mathrm{g}(\mathrm{x}))\times \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\mathrm{g}(\mathrm{x})$.
• $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{u}}\times \frac{\mathrm{d}\mathrm{u}}{\mathrm{d}\mathrm{x}}$.

Calculation:

We have y = ${\cos }^{-1}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)$.

Let x = tan² z.

∴ y = ${\cos }^{-1}\left(\frac{1-{\tan }^2\mathrm{z}}{1+{\tan }^2\mathrm{z}}\right)$ = cos-1 (cos 2z) = 2z.

Now, differentiating w.r.t. z, we get:

$\frac{\mathrm{d}\mathrm{x}}{\mathrm{d}\mathrm{z}}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{z}}({\tan }^2\mathrm{z})=2\tan \mathrm{z}{\sec }^2\mathrm{z}$

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{z}}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{z}}(2\mathrm{z})=2$

Using the chain rule of derivatives, we get:

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{z}}\times \frac{\mathrm{d}\mathrm{z}}{\mathrm{d}\mathrm{x}}=\frac{2}{2\tan \mathrm{z}{\sec }^2\mathrm{z}}=\frac{1}{\tan \mathrm{z}(1+{\tan }^2\mathrm{z})}=\frac{1}{\sqrt{\mathrm{x}}(1+\mathrm{x})}$.

Concept-

sin 2x = 2sin x cos x.

Calculation-

As sin x = sin 2x

⇒ sin 2x - sin x = 0

⇒ 2 sin x cos x - sin x = 0

⇒ sin x (2cos x - 1) = 0

So either sin x = 0 , in interval [0, 2π] when x = 0, π, 2π 

or 2cos x -1 = 0, i.e cos x = $\frac{1}{2}$ in interval [0,2π] when x = $\frac{\pi }{3},\frac{5\pi }{3}$

∴ total values of x in interval [0, 2π] is 5.

Concept used:

Trigonometry formula

sin 2x = 2sin x cos x

Calculation:

y = $\frac{\left(\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}-\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}\right)}{\mathrm{s}\mathrm{i}\mathrm{n}2\mathrm{x}}$

y = $\frac{\left(\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}-\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}\right)}{2.\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}.\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}}$

⇒ y = $\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}{2.\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}.\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}}-\frac{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}}{2.\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}.\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}}$

⇒ y = $\left(\frac{1}{2\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}}\right)-\left(\frac{1}{2\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}\right)$

⇒ y = $\frac{1}{2}$ (secx - cosecx)

Differentiate both sides, we get

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{1}{2}\left[\frac{\mathrm{d}\left(\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{x}\right)}{\mathrm{d}\mathrm{x}}-\frac{\mathrm{d}\left(\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{x}\right)}{\mathrm{d}\mathrm{x}}\right]$

⇒ $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{1}{2}\left(\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{x}.\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{x}+\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{x}.\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}\right)$

Concept:

${\tan }^{-1}\mathrm{x}+{\tan }^{-1}\mathrm{y}={\tan }^{-1}\left[\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{x}\mathrm{y}}\right]$

$\frac{\mathrm{d}({\tan }^{-1}\mathrm{x})}{\mathrm{d}\mathrm{x}}=\frac{1}{1+{\mathrm{x}}^2}$

Calculation:

Given: y = ${\tan }^{-1}\left[\frac{8\mathrm{x}}{1-15{\mathrm{x}}^2}\right]$

$\mathrm{y}={\tan }^{-1}\left[\frac{5\mathrm{x}+3\mathrm{x}}{1-5\mathrm{x}\cdot 3\mathrm{x}}\right]$

As we know that, ${\tan }^{-1}\mathrm{x}+{\tan }^{-1}\mathrm{y}={\tan }^{-1}\left[\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{x}\mathrm{y}}\right]$

So, $\mathrm{y}={\tan }^{-1}\left[\frac{5\mathrm{x}+3\mathrm{x}}{1-5\mathrm{x}\cdot 3\mathrm{x}}\right]$= tan-1 5x + tan-1 3x

Differentiating with respect to x, we get

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}({\tan }^{-1}5\mathrm{x})}{\mathrm{d}\mathrm{x}}+\frac{\mathrm{d}({\tan }^{-1}3\mathrm{x})}{\mathrm{d}\mathrm{x}}$

$=\frac{5}{1+(5\mathrm{x}{)}^2}+\frac{3}{1+(3\mathrm{x}{)}^2}$

$=\frac{5}{1+25{\mathrm{x}}^2}+\frac{3}{1+9{\mathrm{x}}^2}$

Concept:

Derivatives of Trigonometric Functions:

$$\begin{gathered} \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\sin \mathrm{x}=\cos \mathrm{x}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\cos \mathrm{x}=-\sin \mathrm{x} \\ \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\tan \mathrm{x}={\sec }^2\mathrm{x}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\cot \mathrm{x}=-{\csc }^2\mathrm{x} \\ \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\sec \mathrm{x}=\tan \mathrm{x}\sec \mathrm{x}\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\csc \mathrm{x}=-\cot \mathrm{x}\csc \mathrm{x} \end{gathered}$$

Chain Rule of Derivatives:

• $\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\mathrm{f}(\mathrm{g}(\mathrm{x}))=\frac{\mathrm{d}}{\mathrm{d}\mathrm{g}(\mathrm{x})}\mathrm{f}(\mathrm{g}(\mathrm{x}))\times \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\mathrm{g}(\mathrm{x})$.
• $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{u}}\times \frac{\mathrm{d}\mathrm{u}}{\mathrm{d}\mathrm{x}}$.

Calculation:

We have y = sin (cos2 x2)

Differentiating w.r.t. x, we get:

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}$ = [cos (cos2 x2)] × (2 cos x2 (-sin x2)) (2x)

= -4xcos (cos2 x2) cos x2 sin x2

Concept:

$\frac{\mathrm{d}(\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{x})}{\mathrm{d}\mathrm{x}}=\mathrm{s}\mathrm{e}{\mathrm{c}}^2\mathrm{x}$

$\frac{\mathrm{d}(\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{x})}{\mathrm{d}\mathrm{x}}=\frac{1}{\mathrm{x}}$

$\frac{\mathrm{d}(\mathrm{x})}{\mathrm{d}\mathrm{x}}=1$

fog(x) = f{g(x)}

Calculation:

f(x) = log x + 3x - 10 and g(x) = tanx

fog(x) = f{g(x)} = log(tanx) + 3tanx - 10

⇒ fog'(x) = $\frac{\mathrm{d}[\mathrm{l}\mathrm{o}\mathrm{g}(\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{x})]}{\mathrm{d}\mathrm{x}}+3\frac{\mathrm{d}(\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{x})}{\mathrm{d}\mathrm{x}}-\frac{\mathrm{d}(10)}{\mathrm{d}\mathrm{x}}$

⇒ fog'(x) = $\frac{1}{\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{x}}\frac{\mathrm{d}(\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{x})}{\mathrm{d}\mathrm{x}}+3\mathrm{s}\mathrm{e}{\mathrm{c}}^2\mathrm{x}$

⇒ fog'(x) = $\frac{1}{\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{x}}\times \mathrm{s}\mathrm{e}{\mathrm{c}}^2\mathrm{x}+3\mathrm{s}\mathrm{e}{\mathrm{c}}^2\mathrm{x}$

⇒ fog'(x) = $\frac{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}}\times \frac{1}{\mathrm{c}\mathrm{o}{\mathrm{s}}^2\mathrm{x}}+3\mathrm{s}\mathrm{e}{\mathrm{c}}^2\mathrm{x}$

⇒ fog'(x) = $\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{x}.\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{x}+3\mathrm{s}\mathrm{e}{\mathrm{c}}^2\mathrm{x}$

⇒ fog'(x) = $\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{x}(\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{x}+3\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{x})$