Elementary Properties and Identities MCQ Quiz in मल्याळम - Objective Question with Answer for Elementary Properties and Identities - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept - Use trigonometric identities.

\(sin3A=3sinA - 4 sin^3A\)

Solution -   Let 

  \(sin^{-1}\frac{1}{5}= A \Rightarrow sinA = \frac{1}{5} \)

Now \( sin\left [ 3sin^{-1}\left ( \frac{1}{5} \right ) \right ] = sin\left ( 3A \right ) \)

so use the formula - 

\( sin3A = 3sinA - 4sin^{3}A \)

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

\( = \frac{3}{5} -\frac{4}{125}\)

\( = \frac{75-4}{125} \)

\( = \frac{71}{125} \)

Calculation:

Let sin^-1 x = y. Then x = sin y

Since, \(- 1 \le x \le 0,\) 

therefore  \(\frac{{ - \pi }}{2} \le {\sin ^{ - 1}}x \le 0\) and so \(\frac{{ - \pi }}{2}\)\(\le y \le 0\)

We have,

cos y = \(\sqrt {1 - {{\sin }^2}y}\)

 \(\Rightarrow \cos y = \sqrt {1 - {x^2}} ,\) for \(0 \le y \le \pi\)       ....(i)

Now, \(- \frac{\pi }{2} \le y \le 0 \Rightarrow \frac{\pi }{2} \ge - y \ge 0\)      [From (i)]       

\(\Rightarrow \cos ( - y) = \sqrt {1 - {x^2}}\)

\(\Rightarrow - y = {\cos ^{ - 1}}\sqrt {1 - {x^2}}\)

\( \Rightarrow y = - {\cos ^{ - 1}}\sqrt {1 - {x^2}} \)

Concept:

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

Calculation:

Given: tan-1 a + tan-1 b = π/4

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

\( ⇒ {\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left[ {\frac{{a + b}}{{1 - ab}}} \right]\)

⇒ tan^-1 [a + b / 1 - ab] = π/4

⇒ [a + b / 1 - ab] = tan (π/4) = 1

⇒ a + b = 1 - ab

⇒ a + b + ab = 1

Concept:

2 tan-1 x = tan-1 (2x / 1- x2), - 1 ≤ x ≤ 1

Calculation:

Given: 2 tan-1 (cos x) = tan-1 (2 cosec x)

As we know that, 2 tan-1 x = tan-1 (2x / 1- x2), - 1 ≤ x ≤ 1

⇒ 2 tan-1 (cos x) = tan^-1 (2 cos x / 1 - cos² x)

As we know that, sin² x + cos² x = 1

⇒ 2 tan-1 (cos x) = tan-1 (2 cos x / sin² x)

⇒ tan-1 (2 cos x / sin2 x) = tan-1 (2 cosec x) = tan^-1 (2 / sin x)

⇒ 2 cos x / sin2 x = 2 / sin x

⇒ cos x sin x - sin² x = 0

⇒ sin x (cos x - sin x) = 0

⇒ sin x = 0 or cos x - sin x = 0

⇒ x = 0 or π/4

As we can see that for x = 0 the given equation does not exist

Hence, x = π/4 is the only solution for the given equation.

Concept -

Use trigonometric identities

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

Solution -

The given question is also written in the form

\(cot^{-1}\left ( \frac{1}{2} \right )=\frac{\pi }{2}-sin^{-1}x\)

Now use the identity 

\(cot^{-1}\left ( \frac{1}{2} \right )=cos^{-1}x\)............(i) 

Let \(cot^{-1}\left ( \frac{1}{2} \right )=A\Rightarrow \frac{1}{2}=cotA\)

\(\frac{cosA}{sinA}=\frac{1}{2}\)

\(\frac{cos^2A}{sin^2A}=\frac{1}{4}\)

\(4cos^2A=sin^2A\)

\(4cos^2A= 1-cos^2A \)

Now \(cosA = \frac{1}{\sqrt{5}}\) then 

\(A = cos^{-1}\frac{1}{\sqrt{5}}\)

put the value of A we get 

\(cot^{-1}\left ( \frac{1}{2} \right )=cos^{-1}\frac{1}{\sqrt{5}}\)

put this value in equation (i) we get 

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

Now taking cos of both side we get the value of x

\(x=\frac{1}{\sqrt{5}}\)

So the final answer is \(x=\frac{1}{\sqrt{5}}\) hence option 2 is correct.

Concept:

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

Calculation:

\(sin\left(sin^{-1}\frac{1}{5}+cos^{-1}x\right)=1 \)

⇒ \(sin^{-1}\frac{1}{5}+cos^{-1}x=sin^{-1}(1) \)

⇒  \(sin^{-1}\frac{1}{5}+cos^{-1}x=\frac{\pi}{2}\)

Since \(sin^{-1}x+cos^{-1}x=\frac{\pi}{2}\)

⇒ x = \(\frac{1}{5}\)

∴ The value of x is \(\frac{1}{5}\)

Concept:

tan-1 (1/x) = cot x, x > 0

tan-1 x + tan-1 y = tan-1 (x + y) / (1 - xy), xy < 1

Calculation

Given: tan-1 (2x / 1 - x2) + cot-1 (1 - x2 / 2x) = π/3 such that -1 < x < 1

As we know that, tan-1 (1/x) = cot x, x > 0

⇒ cot-1 (1 - x2 / 2x) = tan-1 (2x / 1 - x2)

⇒ 2 tan-1 (2x / 1 - x2) = π/3

⇒ tan-1 (2x / 1 - x2) = π/6

⇒ 2x / 1 - x² = 1/√3

⇒ x² + 2√3 x - 1 = 0

By comparing the above equation with ax² + bx + c = 0, we get

⇒ a = 1, b = 2√3 and c = -1

By using the formula x = [-b ± √(b² - 4ac)] / 2a, we get

⇒ x = 2 - √3 or - (2 + √3)

But as it is given that -1 < x < 1 ⇒ x ≠ - (2 + √3)

Hence, x = 2 - √3

Calculation: 

Given

\(\sin\rm(\tan^{-1}\frac{4}{5}+\tan^{-1}\frac{4}{3}+\tan^{-1}\frac{1}{9}-\tan^{-1}\frac{1}{7})\)

We know that \(\tan^{-1}a +\tan^{-1}b =\tan^{-1} \left( \frac{a+b}{1-ab} \right)\) and \(\tan^{-1}a -\tan^{-1}b =\tan^{-1} \left( \frac{a-b}{1+ab} \right)\)

\(\sin\rm(\tan^{-1}\frac{4}{5}+\tan^{-1}\frac{4}{3}+\tan^{-1}\frac{1}{9}-\tan^{-1}\frac{1}{7})\) = \(\sin\rm\Big[(\tan^{-1}\frac{4}{3}+\tan^{-1}\frac{1}{9})+ (\tan^{-1}\frac{4}{5} -\tan^{-1}\frac{1}{7})\big]\)

= \(\sin\rm\Big[\tan^{-1} \left( \frac{\frac{4}{3} + \frac{1}{9}}{1-\frac{4}{27}} \right)+ \tan^{-1}\Big(\frac{\frac{4}{5}-\frac{1}{7}}{1+\frac{4}{35}}\Big)\Big]\)

= \(\sin\rm(\tan^{-1} \left( \frac{39/27}{23/27} \right) + \tan^{-1} \left( \frac{23/35}{39/35} \right) )\)

= \(\sin\rm(\tan^{-1} \left( \frac{39}{23} \right) + \tan^{-1} \left( \frac{23}{39} \right) )\)

= \(\sin(\tan^{-1} \left( \frac{39/23+23/39}{1-1} \right) )\)

= \(\sin(\tan^{-1} \left( \frac{2050/897}{0} \right) )\)

= \(\sin(\tan^{-1} \infty )\) = sin(π/2) = 1

∴ \(\sin\rm(\tan^{-1}\frac{4}{5}+\tan^{-1}\frac{4}{3}+\tan^{-1}\frac{1}{9}-\tan^{-1}\frac{1}{7})\) = 1

The correct answer is option (4).

Concept:

\(\rm 2\tan^{-1}{x} = \tan^{-1}{2x\over1 - x^2}\)

\(\rm \cot^{-1}{x}= {π\over2} -\tan^{-1}{x}\)

Calculation:

Let S = \(\tan^{-1}{3\over4} + 2\cot^{-1}{1\over3}\)

S = \(\tan^{-1}{3\over4} + 2\left({π\over2}- \tan^{-1}{1\over3}\right)\)           [∵ \(\rm \cot^{-1}{x}= {π\over2} -\tan^{-1}{x}\)]

S = \(\tan^{-1}{3\over4} + π- 2\tan^{-1}{1\over3}\)

S = \( π +\tan^{-1}{3\over4} - \tan^{-1}{2\times{1\over3}\over1 - ({1\over3})^2}\)

S = \( π +\tan^{-1}{3\over4} - \tan^{-1}{3\over4}\)

S = π 

Concept:

If sin θ = x then θ = sin^-1 x

sin-1 (sin θ) = θ, ∀ θ ∈ [-π/2, π/2]

sin-1 x + cos^-1 x = π / 2, where x ∈ [-1, 1]

Calculation:

Given: \(\sin \left( {{{\sin }^{ - 1}}\frac{1}{5} + {{\cos }^{ - 1}}x} \right) = 1\)

As we know that, if sin θ = x then θ = sin-1 x

⇒ sin^-1 (1/5) + cos^-1 x = sin^-1 (1)

As we know that, sin π/2 = 1

⇒ sin-1 (1/5) + cos-1 x = sin-1 (sin π/2)

As we know that, sin-1 (sin θ) = θ, ∀ θ ∈ [-π/2, π/2]

⇒ sin-1 (1/5) + cos-1 x = π/2

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

Hence, x = 1/5