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

Concept:

\(\rm \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left[ \frac{x+y}{1-xy} \right ]\)

\(\rm \frac{d(\tan^{-1} x)}{dx}= \frac{1}{1+x^2}\)

Calculation:

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

\(\rm y=\rm \tan^{-1}\left[ \frac{5x+3x}{1-5x\cdot 3x} \right ]\)

As we know that, \(\rm \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left[ \frac{x+y}{1-xy} \right ]\)

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

Differentiating with respect to x, we get

\(\rm \frac{dy}{dx}=\frac{d(\tan^{-1} 5x)}{dx}+\frac{d(\tan^{-1} 3x)}{dx}\)

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

\(=\rm \frac{5}{1+25x^2}+\frac{3}{1+9x^2}\)

\( y = \sin^2 \left( \cot^{-1} \sqrt{ \dfrac{1 + x}{1 - x} } \right) \)

Let, \( \cot{\theta} = \sqrt{ \dfrac{1 + x}{1 - x} } \)

\( \cot^2{\theta} = \dfrac{1 + x}{1 - x} \)

\( \Rightarrow x = \dfrac{\cot^2{\theta} - 1}{\cot^2{\theta} + 1} = \dfrac{\cos^2{\theta} - \sin^2{\theta}}{\cos^2{\theta} + \sin^2{\theta}} = \cos{2\theta} \)

Now,

\( y = \sin^2{\theta} \)

\( \Rightarrow \dfrac{dy}{dx} = \dfrac{dy}{d(\theta)} \times \dfrac{d(\theta)}{dx} \)

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

\( \Rightarrow \dfrac{dy}{dx} = \dfrac{-1}{2} \)

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:

\({\cot ^{ - 1}}{\rm{x}} = {\rm{\;}}{\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right)\)

tan (tan^-1 x) = x

Calculation:

Given:

y = tan (cot⁻¹ x)                                      

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

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

Differentiating with respect to x, we get

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {\rm{\;}} - \frac{1}{{{{\rm{x}}^2}}}\)

At x = 1

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \; - \frac{1}{1} = \; - 1\)

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:

? = \(\frac{ \sin 33^\circ \cos 57^\circ + \sec 62^\circ \sin 28^\circ + \cos 33^\circ \sin 57^\circ + \rm cosec 62^\circ \cos 28^\circ}{\tan 15^\circ \tan 35^\circ \tan 60^\circ \tan 55^\circ \tan 75^\circ}\)

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

∴ \(\frac{ \sin 33^\circ \cos 57^\circ + \sec 62^\circ \sin 28^\circ + \cos 33^\circ \sin 57^\circ + \rm cosec 62^\circ \cos 28^\circ}{\tan 15^\circ \tan 35^\circ \tan 60^\circ \tan 55^\circ \tan 75^\circ}\) = √3

Concept:

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

• Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{\;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)
• Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right)} \right] = {\rm{\;f'}}\left( {\rm{x}} \right){\rm{\;g}}\left( {\rm{x}} \right) + {\rm{f}}\left( {\rm{x}} \right){\rm{\;g'}}\left( {\rm{x}} \right)\)

Calculation:

We have to find the value of \(\rm \frac{d(\sin xt)}{dt}\)

\(\rm \frac{d(\sin xt)}{dt} = \frac{d(\sin xt)}{d(xt)} \times \frac{d(xt)}{dt}\\=\cos xt \times x\\=x \cos xt\)

Concept:

Derivatives of Trigonometric Functions:

\(\rm \frac{d}{dx}\sin x=\cos x\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}\cos x=-\sin x\\ \frac{d}{dx}\tan x=\sec^2x\ \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}\cot x=-\csc^2 x\\ \frac{d}{dx}\sec x=\tan x\sec x\ \ \ \ \frac{d}{dx}\csc x=-\cot x\csc x\)

Trigonometric Formulae:

\(\rm \sin 2x = \frac{2\tan x}{1+\tan^2x}\)

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

\(\rm \tan 2x = \frac{2\tan x}{1-\tan^2x}\)

Chain Rule of Derivatives:

• \(\rm \frac{d}{dx}f(g(x))=\frac{d}{d\ g(x)}f(g(x))\times \frac{d}{dx}g(x)\).
• \(\rm \frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}\).

Calculation:

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

Let x = tan² z.

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

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

\(\rm\frac{dx}{dz}= \frac{d}{dz}\left(\tan^2z\right)=2\tan z\sec^2z\)

\(\rm\frac{dy}{dz}= \frac{d}{dz}(2z)=2\)

Using the chain rule of derivatives, we get:

\(\rm\frac{dy}{dx}= \frac{dy}{dz}\times\frac{dz}{dx}=\frac{2}{2\tan z\sec^2z}=\frac{1}{\tan z(1+\tan^2z)}=\frac{1}{\sqrt x(1+x)}\).

Concept used:

Trigonometry formula

sin 2x = 2sin x cos x

Calculation:

y = \(\rm {\rm{\;}}\frac{{\left( {sinx - cosx} \right)}}{{sin2x}}\)

y = \(\rm \frac{{\left( {sinx - cosx} \right)}}{{2.sinx.cosx}}\)

⇒ y = \(\rm \frac{{sinx}}{{2.sinx.cosx}} - \;\frac{{cosx}}{{2.sinx.cosx}}\)

⇒ y = \(\rm \left( {\frac{1}{{2cosx}}} \right)\; - \left( {\frac{1}{{2sinx}}} \right)\)

⇒ y = \(\frac{1}{2}\) (secx - cosecx)

Differentiate both sides, we get

⇒ \(\rm \frac{{dy}}{{dx}} = \frac{1}{2}\left[ {\frac{{d\left( {secx} \right)}}{{dx}} - \frac{{d\left( {cosecx} \right)}}{{dx}}} \right]\)

⇒ \(\rm \frac{{dy}}{{dx}} = \frac{1}{2}\left( {secx.tanx + cosecx.cotx} \right)\)

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 = \( {1} \over {2}\) in interval [0,2π] when x = \(\frac{ π } {3} ,\frac { 5π } {3}\)

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

Concept:

\(\rm \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left[ \frac{x+y}{1-xy} \right ]\)

\(\rm \frac{d(\tan^{-1} x)}{dx}= \frac{1}{1+x^2}\)

Calculation:

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

\(\rm y=\rm \tan^{-1}\left[ \frac{5x+3x}{1-5x\cdot 3x} \right ]\)

As we know that, \(\rm \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left[ \frac{x+y}{1-xy} \right ]\)

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

Differentiating with respect to x, we get

\(\rm \frac{dy}{dx}=\frac{d(\tan^{-1} 5x)}{dx}+\frac{d(\tan^{-1} 3x)}{dx}\)

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

\(=\rm \frac{5}{1+25x^2}+\frac{3}{1+9x^2}\)

Concept:

Derivatives of Trigonometric Functions:

\(\rm \frac{d}{dx}\sin x=\cos x\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}\cos x=-\sin x\\ \frac{d}{dx}\tan x=\sec^2x\ \ \ \ \ \ \ \ \ \ \ \frac{d}{dx}\cot x=-\csc^2 x\\ \frac{d}{dx}\sec x=\tan x\sec x\ \ \ \ \frac{d}{dx}\csc x=-\cot x\csc x\)

Chain Rule of Derivatives:

• \(\rm \frac{d}{dx}f(g(x))=\frac{d}{d\ g(x)}f(g(x))× \frac{d}{dx}g(x)\).
• \(\rm \frac{dy}{dx}=\frac{dy}{du}× \frac{du}{dx}\).

Calculation:

We have y = sin (cos2 x2)

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

\(\rm\frac{dy}{dx}\) = [cos (cos2 x2)] × (2 cos x2 (-sin x2)) (2x)

= -4xcos (cos2 x2) cos x2 sin x2

Concept:

\(\rm \frac {d(tanx)} {dx} = sec^2x\)

\(\rm \frac {d(logx)} {dx} = \frac {1} {x}\)

\(\rm \frac {d(x)} {dx} = 1\)

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) = \(\rm \frac {d[log(tanx)]} {dx} + 3\frac {d(tanx)} {dx} - \frac {d(10)} {dx}\)

⇒ fog'(x) = \(\rm \frac {1} {tanx} \frac {d(tanx)} {dx} + 3 sec^2x\)

⇒ fog'(x) = \(\rm \frac {1} {tanx} \times sec^2x + 3 sec^2x\)

⇒ fog'(x) = \(\rm \frac {cosx} {sinx} \times \frac {1} {cos^2x} + 3 sec^2x\)

⇒ fog'(x) = \(\rm cosecx. secx + 3 sec^2x\)

⇒ fog'(x) = \(\rm secx (cosecx + 3secx)\)